链接:

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

题意:

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

思路:

原题求最小的割点.让s不能到t现在将每个点拆成两个,一个入口一个出口,连一个有向边,每两个点之间连一个无向边.
就变成里求最小割,根据最大流最小割,跑最大流即可.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;

const int MAXN = 200+10;
const int INF = 1e9;

struct Edge
{
    int from, to, cap;
};
vector<Edge> edges;
vector<int> G[MAXN*4];
int Dis[MAXN*4];
int n, m, s, t;

void AddEdge(int from, int to, int cap)
{
    edges.push_back(Edge{from, to, cap});
    edges.push_back(Edge{to, from, 0});
    G[from].push_back(edges.size()-2);
    G[to].push_back(edges.size()-1);
}

bool Bfs()
{
    memset(Dis, -1, sizeof(Dis));
    queue<int> que;
    que.push(s);
    Dis[s] = 0;
    while (!que.empty())
    {
        int u = que.front();
        que.pop();
//        cout << u << endl;
        for (int i = 0;i < G[u].size();i++)
        {
            Edge &e = edges[G[u][i]];
            if (e.cap > 0 && Dis[e.to] == -1)
            {
                Dis[e.to] = Dis[u]+1;
                que.push(e.to);
            }
        }
    }
    return Dis[t] != -1;
}

int Dfs(int u, int flow)
{
    if (u == t)
        return flow;
    int res = 0;
    for (int i = 0;i < G[u].size();i++)
    {
        Edge &e = edges[G[u][i]];
        if (e.cap > 0 && Dis[u]+1 == Dis[e.to])
        {
            int tmp = Dfs(e.to, min(flow, e.cap));
//            cout << "flow:" << e.from << ' ' << e.to << ' ' << tmp << endl;
            e.cap -= tmp;
            flow -= tmp;
            edges[G[u][i]^1].cap += tmp;
            res += tmp;
            if (flow == 0)
                break;
        }
    }
    if (res == 0)
        Dis[u] = -1;
    return res;
}

int MaxFlow()
{
    int res = 0;
    while (Bfs())
    {
        res += Dfs(s, INF);
//        cout << res << endl;
    }
    return res;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    while (cin >> n >> m)
    {
        cin >> s >> t;
        for (int i = 0;i <= n*2;i++)
            G[i].clear();
        edges.clear();
        s = s*2-1;
        t = t*2;
        int w;
        for (int i = 1;i <= n;i++)
        {
            cin >> w;
            AddEdge(i*2-1, i*2, w);
        }
        int u, v;
        for (int i = 1;i <= m;i++)
        {
            cin >> u >> v;
            AddEdge(u*2, v*2-1, INF);
            AddEdge(v*2, u*2-1, INF);
        }
        LL res = MaxFlow();
        cout << res << endl;
    }

    return 0;
}

原文地址：https://www.cnblogs.com/YDDDD/p/11331155.html