poj 3469 网络流最小割

http://poj.org/problem?id=3469

Dual Core CPU

Time Limit: 15000MS Memory Limit: 131072KTotal Submissions: 15941 Accepted: 6926Case 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 11 102 1010 32 3 1000

Sample Output

13

Source

POJ Monthly--2007.11.25, Zhou Dong

11607511lanjiangzhou3469Accepted20220K3500MSC++3692B2013-05-17 20:24:25

#include <iostream>#include<cstdio>#include<cstring>using namespace std;const int INF = 100000000;const int nmax = 20000+100;const int mmax = 1000000;struct EDGE{    int u,v,cap,flow;    int next;    EDGE(int a=0,int b=0,int c=0,int d=0):        u(a),v(b),cap(c),flow(d){}};struct EDGEList{    int start[nmax];    int last[nmax];    int t;    EDGE arc[mmax];    void clear(){//清除操作        t=0;        for(int i=0;i<nmax;i++){            last[i]=-1;        }    }    void Push_back(EDGE edge){//追加操作        edge.next=-1; arc[t]=edge;        if(last[edge.u]!=-1) arc[last[edge.u]].next=t;        else start[edge.u]=t;        last[edge.u]=t;  t++;    }    void add_arc(EDGE edge){//创建双向弧        Push_back(edge);        Push_back(EDGE(edge.v,edge.u,edge.cap));    }}net;int q[2][nmax];//数组模拟滚动数组int q1[2],q2[2],qnow;//模拟队列void push_queue(int a){//入队列    q[qnow][q2[qnow]++]=a;}int pop_queue(){//出队列    return q[qnow^1][q1[qnow^1]++];}void switch_queue(){//滚动数组    qnow^=1;    q1[qnow]=0;    q2[qnow]=0;}bool empty_queue(){//判断队列是否为空    return q1[qnow^1]>=q2[qnow^1];}int size_queue(){//队列大小    return q2[qnow^1]-q1[qnow^1];}int n,m;int dis[nmax];//层次网络（距离标号）int path[nmax],deep;//路径int cur[nmax];bool Bfs(){//Bfs构建层次网络    int l,u,v;    for(int i=0;i<nmax;i++){        dis[i]=-1; //初始化    }    dis[0]=0; qnow=0;    switch_queue();    push_queue(0);    switch_queue();    while(!empty_queue()){        l=size_queue();        while(l--){            u=pop_queue();//取出队列结点            for(int i=net.start[u];i!=-1;i=net.arc[i].next){                v=net.arc[i].v;                if(dis[v]==-1&&net.arc[i].cap>net.arc[i].flow){                    push_queue(v);                    dis[v]=dis[u]+1;                    if(v==n)  return true;//汇点在层次网络中                }            }        }        switch_queue();//滚动队列    }    return false;}int Dinic(){//Dinic求最大流    int u,neck,pos,res,i;    int maxflow=0;    while(Bfs()){        memcpy(cur,net.start,sizeof(cur));        deep=0;        u=0;        while(1){//最短路径增广            if(u==n){                neck=INF;                for( i=0;i<deep;i++){                    res=net.arc[path[i]].cap-net.arc[path[i]].flow;                    if(res<neck){                        neck=res;                        pos=i;                    }                }                maxflow+=neck;                for( i=0;i<deep;i++){                    net.arc[path[i]].flow+=neck;                    net.arc[path[i]^1].flow-=neck;                }                deep=pos;                u=net.arc[path[deep]].u;            }            //在层次网络中进行增广            for(i=cur[u];i!=-1;i=net.arc[i].next){                if(net.arc[i].cap>net.arc[i].flow&&dis[u]+1==dis[net.arc[i].v])                break;            }            cur[u]=i;            if(i!=-1){                path[deep++]=i;                u=net.arc[i].v;            }            else {                if(deep==0) break;                dis[u]=-1;                u=net.arc[path[--deep]].u;            }        }    }    return maxflow;//返回最大流}int main(){    int a,b,w;    scanf("%d%d",&n,&m);    net.clear();//初始化    for(int i=1;i<=n;i++){        scanf("%d%d",&a,&b);        net.add_arc(EDGE(0,i,a));        net.add_arc(EDGE(i,n+1,b));    }    for(int i=0;i<m;i++){        scanf("%d%d%d",&a,&b,&w);        net.add_arc(EDGE(a,b,w));    }    n++;    printf("%d\n",Dinic());    return 0;}

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