POJ

Dual Core CPU

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

题意:有N个任务需要分配给两个核心A和B，并且需要花费Ai或者Bi，最后有一些任务之间需要花费Wi传递数据如果他们不再同一个核心上运行；问解决所有任务最小花费是多少。

思路:“用最小的费用将对象分成两个集合的问题，常常可以转换成最小割后解决。考虑把N个模块按照在哪个核心上解决分成两个集合。再求最大流”(《挑战程序设计竞赛》)。以模块为顶点。s向每个模块连一条容量为Ai的线，每个模块向t连一条容量为Bi的线，最后考虑模块之间的数据传输费用。

#include <iostream>#include <fstream>#include <cstdio>#include <cstring>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <cmath>#include <algorithm>#include <functional>#define inf 0X3f3f3f3fusing namespace std;typedef long long ll;const int MAXN=1e5+10;const int MAX=200000+10;const double eps=1e-6;int N,M;//dinic//O(|E||V|)int n;struct EDGE{    int v,c,rev;};vector<EDGE>G[MAX];int level[MAX];//定点到原点的距离标号int iter[MAX];//当前弧void init(){    for(int i=1;i<=N+10;i++)        G[i].clear();}void addedge(int u,int v,int cap){    G[u].push_back((EDGE){v,cap,G[v].size()});    G[v].push_back((EDGE){u,0,G[u].size()-1});}//BFS计算从源点出发的距离标号void bfs(int s){    memset(level,-1,sizeof(level));    queue<int>q;    level[s]=0;    q.push(s);    while(q.size()){        int u=q.front();q.pop();        for(int i=0;i<G[u].size();i++){            EDGE &e=G[u][i];            if(e.c>0&&level[e.v]<0){                level[e.v]=level[u]+1;                q.push(e.v);            }        }    }}//dfs寻找增广路int dfs(int v,int t,int f){    if(v==t)    return f;    for(int &i=iter[v];i<G[v].size();i++){        EDGE &e=G[v][i];        if(e.c>0&&level[v]<level[e.v]){            int d=dfs(e.v,t,min(f,e.c));            if(d>0){                e.c-=d;                G[e.v][e.rev].c+=d;                return d;            }        }    }    return 0;}int max_flow(int s,int t){    int flow=0;    while(1){        bfs(s);        if(level[t]<0)  return flow;        memset(iter,0,sizeof(iter));        int f;        while((f=dfs(s,t,inf))>0)            flow+=f;    }}int main(){    #ifdef ONLINE_JUDGE    #else    freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    #endif    cin>>N>>M;    init();    int u,v,w1,w2,s,t;    s=N+1;t=s+1;    for(int i=1;i<=N;i++){        scanf("%d%d",&w1,&w2);        addedge(s,i,w1);        addedge(i,t,w2);    }    for(int i=1;i<=M;i++){        scanf("%d%d%d",&u,&v,&w1);        addedge(u,v,w1);        addedge(v,u,w1);    }    cout<<max_flow(s,t)<<endl;    return 0;}