HDU 4289 Control (网络流,最大流)

HDU 4289 Control (网络流,最大流)

Description

You, the head of Department of Security, recently received a top-secret information that a group of terrorists is planning to transport some WMD(Weapon of Mass Destruction)from one city (the source) to another one (the destination). You know their date, source and destination, and they are using the highway network.
The highway network consists of bidirectional highways, connecting two distinct city. A vehicle can only enter/exit the highway network at cities only.
You may locate some SA (special agents) in some selected cities, so that when the terrorists enter a city under observation (that is, SA is in this city), they would be caught immediately.
It is possible to locate SA in all cities, but since controlling a city with SA may cost your department a certain amount of money, which might vary from city to city, and your budget might not be able to bear the full cost of controlling all cities, you must identify a set of cities, that:
all traffic of the terrorists must pass at least one city of the set.
sum of cost of controlling all cities in the set is minimal.
You may assume that it is always possible to get from source of the terrorists to their destination.

Input

There are several test cases.
The first line of a single test case contains two integer N and M ( 2 <= N <= 200; 1 <= M <= 20000), the number of cities and the number of highways. Cities are numbered from 1 to N.
The second line contains two integer S,D ( 1 <= S,D <= N), the number of the source and the number of the destination.
The following N lines contains costs. Of these lines the ith one contains exactly one integer, the cost of locating SA in the ith city to put it under observation. You may assume that the cost is positive and not exceeding 10 7.
The followingM lines tells you about highway network. Each of these lines contains two integers A and B, indicating a bidirectional highway between A and B.
Please process until EOF (End Of File).

Output

For each test case you should output exactly one line, containing one integer, the sum of cost of your selected set.
See samples for detailed information.

Sample Input

5 6
5 3
5
2
3
4
12
1 5
5 4
2 3
2 4
4 3
2 1

Sample Output

3

Http

HDU:https://vjudge.net/problem/HDU-4289

Source

网络流,最大流,最小割

题目大意

在一个无向图中,每一个点都有一个点权。现在给定起点S和终点T,求割去若干个点后S与T不连通的最小割去点权和。

解决思路

根据最大流最小割定理,求解出最大流即为最小割(理性理解一下,确实如此)
因为本题的权在点上,所以把点i拆成两个,i和i+n,这两个点之间边的容量就是点权。而其他的边的容量都置为无穷大。然后我们以S为源点,T+n为汇点,求解出最大流就是最小割。
这里使用Dinic实现最大流,可以参考这篇文章

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxN=600;
const int maxM=20001*20;
const int inf=2147483647;

class Edge
{
public:
    int u,v,flow;
};

int n,m;
int S,T;
int cnt=-1;
int Head[maxN];
int Next[maxM];
Edge E[maxM];
int depth[maxN];
int Q[maxM];
int cur[maxN];

void Add_Edge(int u,int v,int flow);
bool bfs();
int dfs(int u,int flow);

int main()
{
    while (cin>>n>>m)
    {
        cnt=-1;
        memset(Head,-1,sizeof(Head));
        scanf("%d%d",&S,&T);//源点和汇点
        T=T+n;//这里要把T变成拆点后的后一个点
        for (int i=1;i<=n;i++)
        {
            int cost;
            scanf("%d",&cost);
            Add_Edge(i,i+n,cost);//读入点权,并拆点连边
        }
        for (int i=1;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);//读入边,连边,注意是无向的
            Add_Edge(u+n,v,inf);
            Add_Edge(v+n,u,inf);
        }
        int Ans=0;//求解最大流,最大流就是最小割
        while (bfs())
        {
            for (int i=1;i<=2*n;i++)
                cur[i]=Head[i];
            while (int di=dfs(S,inf))
                Ans+=di;
        }
        printf("%d\n",Ans);
    }
    return 0;
}

void Add_Edge(int u,int v,int flow)
{
    cnt++;
    Next[cnt]=Head[u];
    Head[u]=cnt;
    E[cnt].u=u;
    E[cnt].v=v;
    E[cnt].flow=flow;

    cnt++;
    Next[cnt]=Head[v];
    Head[v]=cnt;
    E[cnt].u=v;
    E[cnt].v=u;
    E[cnt].flow=0;
    return;
}

bool bfs()
{
    memset(depth,-1,sizeof(depth));
    int h=1,t=0;
    Q[1]=S;
    depth[S]=1;
    do
    {
        t++;
        int u=Q[t];
        for (int i=Head[u];i!=-1;i=Next[i])
        {
            int v=E[i].v;
            if ((depth[v]==-1)&&(E[i].flow>0))
            {
                depth[v]=depth[u]+1;
                h++;
                Q[h]=v;
            }
        }
    }
    while (t!=h);
    if (depth[T]==-1)
        return 0;
    return 1;
}

int dfs(int u,int flow)
{
    if (u==T)
        return flow;
    for (int &i=cur[u];i!=-1;i=Next[i])
    {
        int v=E[i].v;
        if ((depth[v]==depth[u]+1)&&(E[i].flow>0))
        {
            int di=dfs(v,min(flow,E[i].flow));
            if (di>0)
            {
                E[i].flow-=di;
                E[i^1].flow+=di;
                return di;
            }
        }
    }
    return 0;
}
posted @ 2017-08-17 16:42 SYCstudio 阅读(...) 评论(...) 编辑 收藏