Description
Consider the two networks shown below. Assuming that data moves around these networks only between directly connected nodes on a peer-to-peer basis, a failure of a single node, 3, in the network on the left would prevent some of the still available nodes from communicating with each other. Nodes 1 and 2 could still communicate with each other as could nodes 4 and 5, but communication between any other pairs of nodes would no longer be possible.
Node 3 is therefore a Single Point of Failure (SPF) for this network. Strictly, an SPF will be defined as any node that, if unavailable, would prevent at least one pair of available nodes from being able to communicate on what was previously a fully connected network. Note that the network on the right has no such node; there is no SPF in the network. At least two machines must fail before there are any pairs of available nodes which cannot communicate.
Node 3 is therefore a Single Point of Failure (SPF) for this network. Strictly, an SPF will be defined as any node that, if unavailable, would prevent at least one pair of available nodes from being able to communicate on what was previously a fully connected network. Note that the network on the right has no such node; there is no SPF in the network. At least two machines must fail before there are any pairs of available nodes which cannot communicate.
Input
The input will contain the description of several networks. A network description will consist of pairs of integers, one pair per line, that identify connected nodes. Ordering of the pairs is irrelevant; 1 2 and 2 1 specify the same connection. All node numbers will range from 1 to 1000. A line containing a single zero ends the list of connected nodes. An empty network description flags the end of the input. Blank lines in the input file should be ignored.
Output
For each network in the input, you will output its number in the file, followed by a list of any SPF nodes that exist.
The first network in the file should be identified as "Network #1", the second as "Network #2", etc. For each SPF node, output a line, formatted as shown in the examples below, that identifies the node and the number of fully connected subnets that remain when that node fails. If the network has no SPF nodes, simply output the text "No SPF nodes" instead of a list of SPF nodes.
The first network in the file should be identified as "Network #1", the second as "Network #2", etc. For each SPF node, output a line, formatted as shown in the examples below, that identifies the node and the number of fully connected subnets that remain when that node fails. If the network has no SPF nodes, simply output the text "No SPF nodes" instead of a list of SPF nodes.
Sample Input
1 2 5 4 3 1 3 2 3 4 3 5 0 1 2 2 3 3 4 4 5 5 1 0 1 2 2 3 3 4 4 6 6 3 2 5 5 1 0 0
Sample Output
Network #1 SPF node 3 leaves 2 subnets Network #2 No SPF nodes Network #3 SPF node 2 leaves 2 subnets SPF node 3 leaves 2 subnets
二、分?br> 用DFS解决问题Q详l算法：(x)割点与桥?/span>
三、代?br>
1
#include<iostream>
2#include<list>
3using namespace std;
4int t;
5int v1, v2;
6list<int> g[1001];
7bool flag;
8int root;
9int counter;
10bool spf[1001];
11int low[1001], lab[1001];
12bool visit[1001];
13void dfs(int u, int fa)
14
{
15
low[u] = lab[u] = counter++;
16 list<int>::iterator it;
17 int counter = 0;
18 for(it = g[u].begin(); it != g[u].end(); it++)
19
{
20 int v = *it;
21 if(!lab[v])
22 {
23 counter++;
24 dfs(v, u);
25 low[u] = min(low[u], low[v]);
26 if((u==root && counter>=2) || (u!=root && low[v]>=lab[u]))
27 spf[u] = flag = true;
28
}
29 else if(v != fa)
30 low[u] = min(low[u], lab[v]);
31 }
32
}
33void find(int u)
34{
35 visit[u] = true;
36 list<int>::iterator it;
37 for(it = g[u].begin(); it != g[u].end(); it++)
38 if(!visit[*it])
39 find(*it);
40}
41int main()
42{
43 t = 1;
44 while(1)
45 {
46 scanf("%d", &v1);
47 if(v1 == 0) break;
48 for(int i=1; i<=1000; i++)
49 g[i].clear();
50 memset(spf, 0, sizeof spf);
51 memset(low, 0, sizeof low);
52 memset(lab, 0, sizeof lab);
53 while(v1 != 0)
54 {
55 scanf("%d", &v2);
56 g[v1].push_back(v2);
57 g[v2].push_back(v1);
58 root = v1;
59 scanf("%d", &v1);
60 }
61 counter = 1;
62 flag = false;
63 dfs(root, -1);
64 printf("Network #%d\n", t++);
65 if(flag)
66 {
67 for(int i=1; i<=1000; i++)
68 {
69 if(!spf[i]) continue;
70 int cnt = 0;
71 memset(visit, 0, sizeof visit);
72 visit[i] = true;
73 list<int>::iterator it;
74 for(it = g[i].begin(); it != g[i].end(); it++)
75 if(!visit[*it])
76 {
77 cnt++;
78 find(*it);
79 }
80 printf(" SPF node %d leaves %d subnets\n", i, cnt);
81 }
82 }
83 else
84 printf(" No SPF nodes\n");
85 printf("\n");
86 }
87}
#include<iostream>2
#include<list>3
using namespace std;4
int t;5
int v1, v2;6
list<int> g[1001];7
bool flag;8
int root;9
int counter;10
bool spf[1001];11
int low[1001], lab[1001];12
bool visit[1001];13
void dfs(int u, int fa)14
{15
low[u] = lab[u] = counter++;16
list<int>::iterator it;17
int counter = 0;18
for(it = g[u].begin(); it != g[u].end(); it++)19
{20
int v = *it;21
if(!lab[v])22
{23
counter++;24
dfs(v, u);25
low[u] = min(low[u], low[v]);26
if((u==root && counter>=2) || (u!=root && low[v]>=lab[u]))27
spf[u] = flag = true;28
}29
else if(v != fa)30
low[u] = min(low[u], lab[v]);31
}32
}33
void find(int u)34
{35
visit[u] = true;36
list<int>::iterator it;37
for(it = g[u].begin(); it != g[u].end(); it++)38
if(!visit[*it])39
find(*it);40
}41
int main()42
{43
t = 1;44
while(1)45
{46
scanf("%d", &v1);47
if(v1 == 0) break;48
for(int i=1; i<=1000; i++)49
g[i].clear();50
memset(spf, 0, sizeof spf);51
memset(low, 0, sizeof low);52
memset(lab, 0, sizeof lab);53
while(v1 != 0)54
{55
scanf("%d", &v2);56
g[v1].push_back(v2);57
g[v2].push_back(v1);58
root = v1;59
scanf("%d", &v1);60
}61
counter = 1;62
flag = false;63
dfs(root, -1);64
printf("Network #%d\n", t++);65
if(flag)66
{67
for(int i=1; i<=1000; i++)68
{69
if(!spf[i]) continue;70
int cnt = 0;71
memset(visit, 0, sizeof visit);72
visit[i] = true;73
list<int>::iterator it;74
for(it = g[i].begin(); it != g[i].end(); it++)75
if(!visit[*it])76
{77
cnt++;78
find(*it);79
}80
printf(" SPF node %d leaves %d subnets\n", i, cnt);81
}82
}83
else84
printf(" No SPF nodes\n");85
printf("\n");86
}87
}]]>
Problem Description
There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.
Input
First line is a single integer T(T<=10), indicating the number of test cases.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
Output
For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.
Sample Input
2 3 2 1 2 10 3 1 15 1 2 2 3 2 2 1 2 100 1 2 2 1
Sample Output
10 25 100 100
二、分?br> 用Tarjan解决的LCA问题Q详l算法：(x)LCA问题?/span>
三、代?/p>
1
#include<iostream>
2#include<list>
3using namespace std;
4struct node
5{
6 int v, d;
7 void init(int vv, int dd) {v = vv; d = dd;};
8};
9int t, n, m, v1, v2, len;
10list<node> g[40001];
11list<node> query[40001];
12int dis[40001];
13int res[201][3];
14int parent[40001];
15bool visit[40001];
16int find(int k)
17{
18 if(parent[k] == k)
19 return k;
20 return parent[k] = find(parent[k]);
21}
22void tarjan(int u)
23{
24 if(visit[u]) return;
25 visit[u] = true;
26 parent[u] = u;
27 list<node>::iterator it = query[u].begin();
28 while(it != query[u].end())
29 {
30 if(visit[it->v])
31 res[it->d][2] = find(it->v);
32 it++;
33 }
34 it = g[u].begin();
35 while(it != g[u].end())
36 {
37 if(!visit[it->v])
38 {
39 dis[it->v] = min(dis[it->v], dis[u] + it->d);
40 tarjan(it->v);
41 parent[it->v] = u;
42 }
43 it++;
44 }
45}
46int main()
47{
48 scanf("%d", &t);
49 while(t--)
50 {
51 scanf("%d%d", &n, &m);
52 for(int i=1; i<=n; i++)
53 g[i].clear();
54 for(int i=1; i<=m; i++)
55 query[i].clear();
56 for(int i=1; i<n; i++)
57 {
58 scanf("%d%d%d", &v1, &v2, &len);
59 node n1; n1.init(v2, len);
60 g[v1].push_back(n1);
61 node n2; n2.init(v1, len);
62 g[v2].push_back(n2);
63 }
64 for(int i=1; i<=m; i++)
65 {
66 scanf("%d%d", &v1, &v2);
67 res[i][0] = v1;
68 res[i][1] = v2;
69 node n1; n1.init(v2, i);
70 query[v1].push_back(n1);
71 node n2; n2.init(v1, i);
72 query[v2].push_back(n2);
73 }
74 memset(visit, 0, sizeof(visit));
75 memset(dis, 0x7f, sizeof(dis));
76 dis[1] = 0;
77 tarjan(1);
78 for(int i=1; i<=m; i++)
79 printf("%d\n", dis[res[i][0]] + dis[res[i][1]] - 2*dis[res[i][2]]);
80 }
81}
]]>
]]>
]]>
]]>
]]>
Description
A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= pmax(u) of power, may consume an amount 0 <= c(u) <= min(s(u),cmax(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= lmax(u,v) of power delivered by u to v. Let Con=Σuc(u) be the power consumed in the net. The problem is to compute the maximum value of Con.
An example is in figure 1. The label x/y of power station u shows that p(u)=x and pmax(u)=y. The label x/y of consumer u shows that c(u)=x and cmax(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and lmax(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6.
An example is in figure 1. The label x/y of power station u shows that p(u)=x and pmax(u)=y. The label x/y of consumer u shows that c(u)=x and cmax(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and lmax(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6.
Input
There are several data sets in the input. Each data set encodes a power network. It starts with four integers: 0 <= n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers), and 0 <= m <= n^2 (power transport lines). Follow m data triplets (u,v)z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of lmax(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of pmax(u). The data set ends with nc doublets (u)z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of cmax(u). All input numbers are integers. Except the (u,v)z triplets and the (u)z doublets, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.
Output
For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. Each result has an integral value and is printed from the beginning of a separate line.
Sample Input
2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20 7 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7 (3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5 (0)5 (1)2 (3)2 (4)1 (5)4
Sample Output
15 6
二、分?br> 增加点n为sQ点n+1为tQ求最大流Q用Push-Relabel法Q具体算法：(x)最大流问题
三、代?/strong>
1#include<iostream>
2using namespace std;
3#define MAXN 202
4int s, t;
5int n, np, nc, m;
6char str[50];
7int c[MAXN][MAXN];
8int f[MAXN][MAXN];
9int e[MAXN];
10int h[MAXN];
11void push(int u, int v)
12{
13 int d = min(e[u], c[u][v] - f[u][v]);
14 f[u][v] += d;
15 f[v][u] = -f[u][v];
16 e[u] -= d;
17 e[v] += d;
18}
19bool relabel(int u)
20{
21 int mh = INT_MAX;
22 for(int i=0; i<n+2; i++)
23 {
24 if(c[u][i] > f[u][i])
25 mh = min(mh, h[i]);
26 }
27 if(mh == INT_MAX)
28 return false; //D留|络中无从u出发的\
29 h[u] = mh + 1;
30 for(int i=0; i<n+2; i++)
31 {
32 if(e[u] == 0) //已无余流Q不需再次push
33 break;
34 if(h[i] == mh && c[u][i] > f[u][i]) //push的条?/span>
35 push(u, i);
36 }
37 return true;
38}
39void init_preflow()
40{
41 memset(h, 0, sizeof(h));
42 memset(e, 0, sizeof(e));
43 h[s] = n+2;
44 for(int i=0; i<n+2; i++)
45 {
46 if(c[s][i] == 0)
47 continue;
48 f[s][i] = c[s][i];
49 f[i][s] = -f[s][i];
50 e[i] = c[s][i];
51 e[s] -= c[s][i];
52 }
53}
54void push_relabel()
55{
56 init_preflow();
57 bool flag = true; //表示是否q有relabel操作
58 while(flag)
59 {
60 flag = false;
61 for(int i=0; i<n; i++)
62 if(e[i] > 0)
63 flag = flag || relabel(i);
64 }
65}
66int main()
67{
68 while(scanf("%d%d%d%d", &n, &np, &nc, &m) != EOF)
69 {
70 s = n; t = n+1;
71 memset(c, 0, sizeof(c));
72 memset(f, 0, sizeof(f));
73 while(m--)
74 {
75 scanf("%s", &str);
76 int u=0, v=0, z=0;
77 sscanf(str, "(%d,%d)%d", &u, &v, &z);
78 c[u][v] = z;
79 }
80 for(int i=0; i<np+nc; i++)
81 {
82 scanf("%s", &str);
83 int u=0, z=0;
84 sscanf(str, "(%d)%d", &u, &z);
85 if(i < np)
86 c[s][u] = z;
87 else if(i >= np && i < np + nc)
88 c[u][t] = z;
89 }
90 push_relabel();
91 printf("%d\n", e[t]);
92 }
93}
#include<iostream>2
using namespace std;3
#define MAXN 2024
int s, t;5
int n, np, nc, m;6
char str[50];7
int c[MAXN][MAXN];8
int f[MAXN][MAXN];9
int e[MAXN];10
int h[MAXN];11
void push(int u, int v)12
{13
int d = min(e[u], c[u][v] - f[u][v]);14
f[u][v] += d;15
f[v][u] = -f[u][v];16
e[u] -= d;17
e[v] += d;18
}19
bool relabel(int u)20
{21
int mh = INT_MAX;22
for(int i=0; i<n+2; i++)23
{24
if(c[u][i] > f[u][i])25
mh = min(mh, h[i]);26
}27
if(mh == INT_MAX)28
return false; //D留|络中无从u出发的\29
h[u] = mh + 1;30
for(int i=0; i<n+2; i++)31
{32
if(e[u] == 0) //已无余流Q不需再次push33
break;34
if(h[i] == mh && c[u][i] > f[u][i]) //push的条?/span>35
push(u, i);36
}37
return true;38
}39
void init_preflow()40
{41
memset(h, 0, sizeof(h));42
memset(e, 0, sizeof(e));43
h[s] = n+2;44
for(int i=0; i<n+2; i++)45
{46
if(c[s][i] == 0)47
continue;48
f[s][i] = c[s][i];49
f[i][s] = -f[s][i];50
e[i] = c[s][i];51
e[s] -= c[s][i];52
}53
}54
void push_relabel()55
{56
init_preflow();57
bool flag = true; //表示是否q有relabel操作58
while(flag)59
{60
flag = false;61
for(int i=0; i<n; i++)62
if(e[i] > 0)63
flag = flag || relabel(i);64
}65
}66
int main()67
{68
while(scanf("%d%d%d%d", &n, &np, &nc, &m) != EOF)69
{70
s = n; t = n+1;71
memset(c, 0, sizeof(c));72
memset(f, 0, sizeof(f));73
while(m--)74
{75
scanf("%s", &str);76
int u=0, v=0, z=0;77
sscanf(str, "(%d,%d)%d", &u, &v, &z);78
c[u][v] = z;79
}80
for(int i=0; i<np+nc; i++)81
{82
scanf("%s", &str);83
int u=0, z=0;84
sscanf(str, "(%d)%d", &u, &z);85
if(i < np)86
c[s][u] = z;87
else if(i >= np && i < np + nc)88
c[u][t] = z;89
}90
push_relabel();91
printf("%d\n", e[t]);92
}93
}]]>
Description
Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.
Input
The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.
Output
For each case, output a single integer, the maximum rate at which water may emptied from the pond.
Sample Input
5 4 1 2 40 1 4 20 2 4 20 2 3 30 3 4 10
Sample Output
50
二、分?br> 其实是以点1为sQ以点m为tQ求最大流Q但是要注意输入的\径可以重?见代?0?Q用Edmonds-Karp法Q具体算法：(x)最大流问题
三、代?br>
1#include<iostream>
2#include<queue>
3using namespace std;
4#define MAXM 201
5int m, n;
6int si, ei, ci;
7int c[MAXM][MAXM];
8int f[MAXM][MAXM];
9int cf[MAXM][MAXM];
10bool visit[MAXM];
11int p[MAXM];
12struct node
13{
14 int v, cf;
15 void set(int vv, int ccf)
16 {
17 v = vv; cf = ccf;
18 }
19};
20int main()
21{
22 while(scanf("%d%d", &n, &m) != EOF)
23 {
24 memset(c, 0, sizeof(c));
25 memset(f, 0, sizeof(f));
26 memset(cf, 0, sizeof(cf));
27 while(n--)
28 {
29 scanf("%d%d%d", &si, &ei, &ci);
30 c[si][ei] += ci;
31 cf[si][ei] = c[si][ei];
32 }
33 bool flag = true; //用于表示是否扑ֈ增广?/span>
34 while(flag)
35 {
36 flag = false;
37 memset(visit, 0, sizeof(visit));
38 queue<node> q;
39 node temp;
40 temp.set(1, INT_MAX);
41 p[1] = 0;
42 q.push(temp); visit[1] = true;
43 while(!q.empty()) //q度优先搜烦(ch)
44 {
45 node temp = q.front(); q.pop();
46 for(int i=1; i<=m; i++)
47 {
48 if(temp.v == i || visit[i] || cf[temp.v][i] == 0)
49 continue;
50 node newNode;
51 newNode.set(i, min(temp.cf, cf[temp.v][i]));
52 p[i] = temp.v;
53 q.push(newNode);
54 visit[i] = true;
55 if(i == m)
56 {
57 flag = true; //扑ֈ增广?/span>
58 break;
59 }
60 }
61 if(flag)
62 break;
63 }
64 if(flag)
65 {
66 int mincf = q.back().cf;
67 int v1 = p[m], v2 = m;
68 while(v1 != 0)
69 {
70 f[v1][v2] += mincf; //修改?/span>
71 f[v2][v1] = -f[v1][v2];
72 cf[v1][v2] = c[v1][v2] - f[v1][v2]; //修改D留定w
73 cf[v2][v1] = c[v2][v1] - f[v2][v1];
74 v2 = v1;
75 v1 = p[v1];
76 }
77 }
78 }
79 int res = 0;
80 for(int i=2; i<=m; i++) //计算最大流
81 res += f[1][i];
82 printf("%d\n", res);
83 }
84}
#include<iostream>2
#include<queue>3
using namespace std;4
#define MAXM 2015
int m, n;6
int si, ei, ci;7
int c[MAXM][MAXM];8
int f[MAXM][MAXM];9
int cf[MAXM][MAXM];10
bool visit[MAXM];11
int p[MAXM];12
struct node13
{14
int v, cf;15
void set(int vv, int ccf)16
{17
v = vv; cf = ccf;18
}19
};20
int main()21
{22
while(scanf("%d%d", &n, &m) != EOF)23
{24
memset(c, 0, sizeof(c));25
memset(f, 0, sizeof(f));26
memset(cf, 0, sizeof(cf));27
while(n--)28
{29
scanf("%d%d%d", &si, &ei, &ci);30
c[si][ei] += ci;31
cf[si][ei] = c[si][ei];32
}33
bool flag = true; //用于表示是否扑ֈ增广?/span>34
while(flag)35
{36
flag = false;37
memset(visit, 0, sizeof(visit));38
queue<node> q;39
node temp;40
temp.set(1, INT_MAX);41
p[1] = 0;42
q.push(temp); visit[1] = true;43
while(!q.empty()) //q度优先搜烦(ch)44
{45
node temp = q.front(); q.pop();46
for(int i=1; i<=m; i++)47
{48
if(temp.v == i || visit[i] || cf[temp.v][i] == 0)49
continue;50
node newNode; 51
newNode.set(i, min(temp.cf, cf[temp.v][i]));52
p[i] = temp.v;53
q.push(newNode);54
visit[i] = true;55
if(i == m)56
{57
flag = true; //扑ֈ增广?/span>58
break;59
}60
}61
if(flag)62
break;63
}64
if(flag)65
{66
int mincf = q.back().cf;67
int v1 = p[m], v2 = m;68
while(v1 != 0)69
{70
f[v1][v2] += mincf; //修改?/span>71
f[v2][v1] = -f[v1][v2];72
cf[v1][v2] = c[v1][v2] - f[v1][v2]; //修改D留定w73
cf[v2][v1] = c[v2][v1] - f[v2][v1];74
v2 = v1;75
v1 = p[v1];76
}77
}78
}79
int res = 0;80
for(int i=2; i<=m; i++) //计算最大流81
res += f[1][i];82
printf("%d\n", res);83
}84
}]]>