hdu 4289 Control（最小割 + 拆点）

http://acm.hdu.edu.cn/showproblem.php?pid=4289

Control

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2247    Accepted Submission(s): 940

Problem 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 ¹ 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.
------------------------------------------------------------
¹ Weapon of Mass Destruction

 

 

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 10⁷.
　　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

 

详细请参考：

http://www.cnblogs.com/liuxin13/p/4719335.html

题目大意：

N个点,每个点都有各自的cost, 然后M 无向条边

要求割去S点到D路线中的点,使之无法从S到D ,而且要求消耗的cost和最小.  

 

 

这是一道网络流的题. 算的是最小割. 根据最大流最小割定理. 可以直接算最大流;

但是这题的的流量限制是在点上的.所以要我们来拆点.

我这题是把i 点的  点首和点尾 分别设为 i 和 i+n;  显然 最后会得到2*n个点

如图：

将点1拆分成两部分分别为点首1和点尾1+n，然后把点首到点尾的流量限制设成 题目要求的cost; ，点1---->（1+n）的花费即为封锁城市1的代价

 

而点与点之间（两座城市之间）的边,要设成正无穷大, 因为边不消耗cost;

然后从S的点首S 跑到 D的点尾 D+n  就可以计算出最小割了.

#include<stdio.h>
#include<algorithm>
#include<queue>
#include<string.h>
#define N 510
#define INF 0x3f3f3f3f
using namespace std;

struct Edge
{
    int u, v, flow, next;
} edge[N * N];

int layer[N], head[N], cnt;

void Init()
{
    memset(head, -1, sizeof(head));
    cnt = 0;
}

void AddEdge(int u, int v, int flow)
{
    edge[cnt].u = u;
    edge[cnt].v = v;
    edge[cnt].flow = flow;
    edge[cnt].next = head[u];
    head[u] = cnt++;

    swap(u, v);

    edge[cnt].u = u;
    edge[cnt].v = v;
    edge[cnt].flow = 0;
    edge[cnt].next = head[u];
    head[u] = cnt++;

}

bool BFS(int Start, int End)
{
    queue<int>Q;
    memset(layer, -1, sizeof(layer));
    Q.push(Start);
    layer[Start] = 1;
    while(!Q.empty())
    {
        int u = Q.front();
        Q.pop();
        if(u == End)
            return true;
        for(int i = head[u] ; i != -1 ; i = edge[i].next)
        {
            int v = edge[i].v;
            if(layer[v] == -1 && edge[i].flow > 0)
            {
                layer[v] = layer[u] + 1;
                Q.push(v);
            }
        }
    }
    return false;
}

int DFS(int u, int Maxflow, int End)
{
    if(u == End)
        return Maxflow;
    int uflow = 0;
    for(int i = head[u] ; i != -1 ; i = edge[i].next)
    {
        int v = edge[i].v;
        if(layer[v] == layer[u] + 1 && edge[i].flow > 0)
        {
            int flow = min(edge[i].flow, Maxflow - uflow);
            flow = DFS(v, flow, End);
            edge[i].flow -= flow;
            edge[i^1].flow += flow;

            uflow += flow;
            if(uflow == Maxflow)
                break;
        }
    }
    if(uflow == 0)
        layer[u] = 0;
    return uflow;
}

int Dinic(int Start, int End)
{
    int Maxflow = 0;
    while(BFS(Start, End))
        Maxflow += DFS(Start, INF, End);
    return Maxflow;
}

int main()
{
    int m, n, s, t;
    while(~scanf("%d%d", &m, &n))
    {
        Init();
        scanf("%d%d", &s, &t);
        int u, v, flow;
        for(int i = 1 ; i <= m ; i++)
        {
            scanf("%d", &flow);
            AddEdge(i, i + m, flow);
        }
        while(n--)
        {
            scanf("%d%d", &u, &v);
            AddEdge(u + m, v, INF);
            AddEdge(v + m, u, INF);
        }
        printf("%d\n", Dinic(s, t + m));
    }
    return 0;
}