poj3469 Dual Core CPU

Dual Core CPU

Time Limit: 15000MS		Memory Limit: 131072K
Total Submissions: 24092		Accepted: 10462
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 A_i and B_i. 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, A_i and B_i.
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

Sample Output

Source

POJ Monthly--2007.11.25, Zhou Dong

#include<cstdio>
#include<iostream>
using namespace std;
const int N=2e4+5;
const int M=1e6+5;
struct edge{int v,next,cap;}e[M];int tot=1,head[N];
int n,m,S,T,dis[N],q[N];
void add(int x,int y,int z){
    e[++tot].v=y;e[tot].cap=z;e[tot].next=head[x];head[x]=tot;
    e[++tot].v=x;e[tot].cap=0;e[tot].next=head[y];head[y]=tot;
}
bool bfs(){
    for(int i=S;i<=T;i++) dis[i]=-1;
    int h=0,t=1;q[t]=S;dis[S]=0;
    while(h!=t){
        int x=q[++h];
        for(int i=head[x];i;i=e[i].next){
            if(e[i].cap&&dis[e[i].v]==-1){
                dis[e[i].v]=dis[x]+1;
                if(e[i].v==T) return 1;
                q[++t]=e[i].v;
            }
        }
    }
    return 0;
}
int dfs(int x,int f){
    if(x==T) return f;
    int used=0,t;
    for(int i=head[x];i;i=e[i].next){
        if(e[i].cap&&dis[e[i].v]==dis[x]+1){
            t=dfs(e[i].v,min(e[i].cap,f));
            e[i].cap-=t;e[i^1].cap+=t;
            used+=t;f-=t;
            if(!f) return used;
        }
    }
    if(!used) dis[x]=-1;
    return used;
}
int dinic(){
    int res=0;
    while(bfs()) res+=dfs(S,2e9);
    return res;
}
void init(){
    scanf("%d%d",&n,&m);S=0;T=n+1;
    for(int i=1,a,b;i<=n;i++) scanf("%d%d",&a,&b),add(S,i,a),add(i,T,b);
    for(int i=1,a,b,w;i<=m;i++) scanf("%d%d%d",&a,&b,&w),add(a,b,w),add(b,a,w);
    printf("%d\n",dinic());
}
int main(){
    init();
    return 0;
}

__EOF__

本文作者：shenben
本文链接：https://www.cnblogs.com/shenben/p/6606399.html
关于博主：评论和私信会在第一时间回复。或者直接私信我。
版权声明：本博客所有文章除特别声明外，均采用 BY-NC-SA 许可协议。转载请注明出处！
声援博主：如果您觉得文章对您有帮助，可以点击文章右下角【推荐】一下。您的鼓励是博主的最大动力！