poj 3469（最大流最小割）_poj monthly--2007.11.25, zhou dong (poj 3469)-优快云博客

本文链接：https://blog.youkuaiyun.com/fp_hzq/article/details/6670079

本文探讨了在双核CPU环境下如何通过合理分配任务模块来最小化总成本的问题，并采用最大流最小割的方法进行求解。

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Dual Core CPU

Time Limit: 15000MS		Memory Limit: 131072K
Total Submissions: 10776		Accepted: 4533
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 wdollars 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

分析：这题转化最大流最小割定理，感觉不是很明白啊。。。

对于源点到每个点i连接一条容量为Ai的有向边，每个点i到汇点连一条容量为Bi的有向边，点i与点j如果有关系，则连一条双向边，容量都为两点不在一起的花费，求最大流即可。。。晕乎乎的~~~

Dinic果然不行啊3000++ms，还有poj数组开小了竟然跑出个TLE。。。

代码：

#include<cstdio>
using namespace std;
const int mm=1000000;
const int mn=22222;
const int oo=1000000000;
int node,src,dest,edge;
int reach[mm],flow[mm],next[mm];
int head[mn],work[mn],dis[mn],q[mn];
inline int min(int a,int b)
{
    return a<b?a:b;
}
inline void prepare(int _node,int _src,int _dest)
{
    node=_node,src=_src,dest=_dest;
    for(int i=0;i<node;++i)head[i]=-1;
    edge=0;
}
inline void addedge(int u,int v,int c1,int c2)
{
    reach[edge]=v,flow[edge]=c1,next[edge]=head[u],head[u]=edge++;
    reach[edge]=u,flow[edge]=c2,next[edge]=head[v],head[v]=edge++;
}
bool Dinic_bfs()
{
    int i,u,v,l,r=0;
    for(i=0;i<node;++i)dis[i]=-1;
    dis[q[r++]=src]=0;
    for(l=0;l<r;++l)
        for(i=head[u=q[l]];i>=0;i=next[i])
            if(flow[i]&&dis[v=reach[i]]<0)
            {
                dis[q[r++]=v]=dis[u]+1;
                if(v==dest)return 1;
            }
    return 0;
}
int Dinic_dfs(int u,int exp)
{
    if(u==dest)return exp;
    for(int &i=work[u],v,tmp;i>=0;i=next[i])
        if(flow[i]&&dis[v=reach[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
        {
            flow[i]-=tmp;
            flow[i^1]+=tmp;
            return tmp;
        }
    return 0;
}
int Dinic_flow()
{
    int i,ret=0,delta;
    while(Dinic_bfs())
    {
        for(i=0;i<node;++i)work[i]=head[i];
        while(delta=Dinic_dfs(src,oo))ret+=delta;
    }
    return ret;
}
int main()
{
    int i,j,n,m,a,b;
    while(scanf("%d%d",&n,&m)!=-1)
    {
        prepare(n+2,0,n+1);
        for(i=1;i<=n;++i)
        {
            scanf("%d%d",&a,&b);
            addedge(src,i,a,0);
            addedge(i,dest,b,0);
        }
        while(m--)
        {
            scanf("%d%d%d",&i,&j,&a);
            addedge(i,j,a,a);
        }
        printf("%d\n",Dinic_flow());
    }
    return 0;
}