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poj3469(最大流最小割问题)

2024-07-04 10:49:27   233人阅读

题目链接：http://poj.org/problem?id=3469

Dual Core CPU
Time Limit: 15000MS Memory Limit: 131072K
Total Submissions: 18147 Accepted: 7833
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

Source

POJ Monthly--2007.11.25, Zhou Dong

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一道典型的最小割问题。

我们先假定两个额外的点，源点s，和汇点t。 我们记图中 s-t 割所对应的包含s的顶点集合为S,包含t集合为T。

对于顶点属于S时所产生的费用，只要从每个模块向t连一条容量为A[i]的边就可以对应起来，对于属于T的，只要从s向每个模块连一条容量为B[i]的边即可，接下对那些a[i]属于S，b[i]属于T时所产生的费用，只要从模块a[i]向模块b[i]连一条容量为w[i]的边就可以对应，对剩余情况同理可得。

code：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int MAX=200010;
const int INF=1<<30;
struct edge{int to,cap,rev;};
vector<edge> G[MAX];
int level[MAX];
int iter[MAX];
void add(int from,int to,int cap)
{
    G[from].push_back((edge){to,cap,G[to].size()});
    G[to].push_back((edge){from,0,G[from].size()-1});
}
void bfs(int s)
{
    memset(level,-1,sizeof(level));
    queue<int> que;
    level[s]=0;
    que.push(s);
    while(!que.empty())
    {
        int v=que.front();
        que.pop();
        for(int i=0;i<G[v].size();i++)
        {
            edge &e=G[v][i];
            if(e.cap>0&& level[e.to]<0)
            {
                level[e.to]=level[v]+1;
                que.push(e.to);
            }
        }
    }
}
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.cap>0&&level[v]<level[e.to])
        {
            int d=dfs(e.to,t,min(f,e.cap));
            if(d>0)
            {
                e.cap-=d;
                G[e.to][e.rev].cap+=d;
                return d;
            }
        }
    }
    return 0;
}
int max_flow(int s,int t)
{
    int flow=0;
    for(;;)
    {
        bfs(s);
        if(level[t]<0) return flow;
        memset(iter,0,sizeof(iter));
        int f;
        while((f=dfs(s,t,INF))>0)
            flow+=f;
    }
}
const int MAXV=20010;
int N,M;
int A[MAXV],B[MAXV];
int a[MAX],b[MAX],w[MAX];
void solve()
{
    int s=N,t=s+1;
    for(int i=0;i<N;i++)
    {
        add(i,t,A[i]);
        add(s,i,B[i]);
    }
    for(int i=0;i<M;i++)
    {
        add(a[i]-1,b[i]-1,w[i]);
        add(b[i]-1,a[i]-1,w[i]);
    }
    cout<<max_flow(s,t)<<endl;
}
int main()
{
    cin>>N>>M;
    for(int i=0;i<N;i++)
    {
        scanf("%d%d",&A[i],&B[i]);
    }
    for(int i=0;i<M;i++)
    {
        scanf("%d%d%d",&a[i],&b[i],&w[i]);
    }
    solve();
    return 0;
}



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