一、題目描述
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.
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、分析
其實(shí)就是以點(diǎn)1為s，以點(diǎn)m為t，求最大流，但是要注意輸入的路徑可以重復(fù)(見(jiàn)代碼30行)，使用Edmonds-Karp算法，具體算法：
最大流問(wèn)題三、代碼
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; //用于表示是否找到增廣路
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()) //廣度優(yōu)先搜索
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; //找到增廣路
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; //修改流
71 f[v2][v1] = -f[v1][v2];
72 cf[v1][v2] = c[v1][v2] - f[v1][v2]; //修改殘留容量
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++) //計(jì)算最大流
81 res += f[1][i];
82 printf("%d\n", res);
83 }
84}
posted on 2009-06-23 19:38
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