POJ 3469 Dual Core CPU （Dinic）

Dual Core CPU

Time Limit: 15000MS		Memory Limit: 131072K
Total Submissions: 15782		Accepted: 6813
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

3 1
1 10
2 10
10 3
2 3 1000

Sample Output

13

Source

POJ Monthly--2007.11.25, Zhou Dong

 

 

题意：现在有n个任务，两个机器A和B，每个任务要么在A上完成，要么在B上完成，而且知道每个任务在A和B机器上完成所需要的费用。然后再给m行，每行 a，b，w三个数字。表示如果a任务和b任务不在同一个机器上工作的话，需要额外花费w。现在要求出完成所有任务最小的花费是多少。

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int INF=0x3f3f3f3f;
const int VM=20010;
const int EM=500000;

struct Edge{
    int u,v;
    int cap,nxt;
}edge[EM];

int head[VM],dep[VM],cnt;
int src,des;

void addedge(int cu,int cv,int cw,int rw){   //无向边的逆边 cap = cw，有向边的cap = 0；
    edge[cnt].u=cu;
    edge[cnt].v=cv;
    edge[cnt].cap=cw;
    edge[cnt].nxt=head[cu];
    head[cu]=cnt++;
    edge[cnt].u=cv;
    edge[cnt].v=cu;
    edge[cnt].cap=rw;
    edge[cnt].nxt=head[cv];
    head[cv]=cnt++;
}

int BFS(){
    int q[VM],front=0,rear=0;
    memset(dep,-1,sizeof(dep));
    q[rear++]=src;
    dep[src]=0;
    while(front!=rear){
        int u=q[front++];
        front=front%VM;
        for(int i=head[u];i!=-1;i=edge[i].nxt){
            int v=edge[i].v;
            if(edge[i].cap>0 && dep[v]==-1){
                dep[v]=dep[u]+1;
                q[rear++]=v;
                rear=rear%VM;
                if(v==des)
                    return 1;
            }
        }
    }
    return 0;
}

int Dinic(){
    int i,top,res=0;
    int stack[VM],cur[VM];
    while(BFS()){
        memcpy(cur,head,sizeof(head));
        int u=src;
        top=0;
        while(1){
            if(u==des){
                int min=INF,loc;
                for(i=0;i<top;i++)
                    if(min>edge[stack[i]].cap){
                        min=edge[stack[i]].cap;
                        loc=i;
                    }
                for(i=0;i<top;i++){
                    edge[stack[i]].cap-=min;
                    edge[stack[i]^1].cap+=min;
                }
                res+=min;
                top=loc;
                u=edge[stack[top]].u;
            }
            for(i=cur[u];i!=-1;cur[u]=i=edge[i].nxt)
                if(edge[i].cap!=0 && dep[u]+1==dep[edge[i].v])
                    break;
            if(cur[u]!=-1){
                stack[top++]=cur[u];
                u=edge[cur[u]].v;
            }else{
                if(top==0)
                    break;
                dep[u]=-1;
                u=edge[stack[--top]].u;
            }
        }
    }
    return res;
}

int main(){

    //freopen("input.txt","r",stdin);

    int n,m;
    while(~scanf("%d%d",&n,&m)){
        memset(head,-1,sizeof(head));
        src=0;
        des=n+1;
        int a,b;
        for(int i=1;i<=n;i++){
            scanf("%d%d",&a,&b);
            addedge(src,i,a,0);
            addedge(i,des,b,0);
        }
        int u,v,w;
        while(m--){
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w,w);
        }
        printf("%d\n",Dinic());
    }
    return 0;
}