Dual Core CPU
| Time Limit: 15000MS |
|
Memory Limit: 131072K |
| Total Submissions: 12960 |
|
Accepted: 5553 |
| Case Time Limit: 5000MS |
Description
As more and more computers are equipped with dual core CPU, SetagLilb, the Chief Technology Officer of TinySoft Corporation, decided to update their famous product - SWODNIW.
The routine consists of N modules, and each of them should run in a certain core. The costs for all the routines to execute on two cores has been estimated. Let's define them as Ai and Bi. Meanwhile, M pairs of modules need to do some data-exchange. If they are running on the same core, then the cost of this action can be ignored. Otherwise, some extra cost are needed. You should arrange wisely to minimize the total cost.
Input
There are two integers in the first line of input data, N and M (1 ≤ N ≤ 20000, 1 ≤ M ≤ 200000) .
The next N lines, each contains two integer, Ai and Bi.
In the following M lines, each contains three integers: a, b, w. The meaning is that if module a and module b don't execute on the same core, you should pay extra w dollars for the data-exchange between them.
Output
Output only one integer, the minimum total cost.
Sample Input
3 1
1 10
2 10
10 3
2 3 1000
Sample Output
13
這題點有20000，邊有200000，所以要寫鏈表或者數組模擬
終于找了個能看懂的了
而且我覺著這代碼特別神有幾個地方
1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4#define nmax 20010
5#define emax 200010
6#define inf 1<<30
7int nn;
8int head[nmax];
9struct node
10
{
11
int v,next,w;
12
};
13struct node edge[emax*8];
14int cnt,n,m,s,t;
15void add(int u,int v,int w)
16{
17 edge[cnt].v=v;
18 edge[cnt].w=w;
19 edge[cnt].next=head[u];
20 head[u]=cnt++;
21 edge[cnt].v=u;
22 edge[cnt].w=0;
23 edge[cnt].next=head[v];
24 head[v]=cnt++;
25}
26int sap()
27{
28 int pre[nmax],cur[nmax],dis[nmax],gap[nmax];
29 int flow,aug,u;
30 int flag;
31 int i;
32 flow=0;
33 aug=inf;
34 for(i=0; i<=nn; i++)
35
{
36 cur[i]=head[i];
37 gap[i]=0;
38 dis[i]=0;
39
}
40 gap[s]=nn;
41 u=s;
42 pre[s]=s;
43 while(dis[s]<nn)
44 {
45 flag=0;
46 for(int &j=cur[u]; j!=-1; j=edge[j].next)
47 {
48 int v=edge[j].v;
49 if(edge[j].w>0&&dis[u]==dis[v]+1)
50 {
51 flag=1;
52 if(edge[j].w<aug)aug=edge[j].w;
53 pre[v]=u;
54 u=v;
55 if (u==t)
56 {
57 flow+=aug;
58 while(u!=s)
59 {
60 u=pre[u];
61 edge[cur[u]].w-=aug;
62 edge[cur[u]^1].w+=aug;
63 //why?解釋偶數異或1為偶數+1，奇數異或1為奇數-1，
64 //顯然我們存的邊是從0開始存的，
65 //所以偶數,偶數+1是殘量網格中的兩條邊(無向邊)
66 }
67 aug=inf;
68 }
69 break;
70 }
71 }
72 if (flag) continue;
73 int mindis=nn;
74 for(int j=head[u];j!=-1;j=edge[j].next)
75 {
76 int v=edge[j].v;
77 if (edge[j].w>0&&dis[v]<mindis)
78 {
79 mindis=dis[v];
80 cur[u]=j;
81 }
82 }
83 if (--gap[dis[u]]==0)//間隙優化
84 {
85 break;
86 }
87 dis[u]=mindis+1;
88 gap[dis[u]]++;
89 u=pre[u];
90 }
91 return flow;
92}
93int main()
94{
95 int a,b,c,i;
96 int ans;
97 while (scanf("%d%d",&n,&m)!=EOF)
98 {
99 s=0;
100 t=n+1;
101 cnt=0;
102 memset(head,-1,sizeof(head));
103 for(i=1; i<=n; i++)
104 {
105 scanf("%d%d",&a,&b);
106 add(s,i,a);
107 add(i,t,b);
108 }
109 for(i=1; i<=m; i++)
110 {
111 scanf("%d%d%d",&a,&b,&c);
112 add(a,b,c);
113 add(b,a,c);
114 }
115 ans=0;
116 nn=n+2;
117 ans=sap();
118 printf("%d\n",ans);
119 }
120 return 0;
121}
122
有兩個需要特別注意的循環撒
那個int j那個很奇怪，怪我語言沒學好