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  • bzoj2718

    二分图匹配

    首先有个定理:最长反链=最小链覆盖

    最小链覆盖可以重复经过点

    所以我们不能直接建图

    那么我们用floyd判断是否相连

    然后建图就行了

    #include<bits/stdc++.h>
    using namespace std;
    const int N = 410, inf = 1e9;
    int rd()
    {
        int x = 0, f = 1; char c = getchar();
        while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
        while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
        return x * f;
    }
    int n, m, k, D, source, sink, cnt = 1, tot;
    int a[N], d[N], head[N], iter[N], Map[N][N];
    struct edge {
        int nxt, to, f;
    } e[N * N];
    bool bfs()
    {
        queue<int> q;
        memset(d, -1, sizeof(d));
        d[source] = 0;
        q.push(source);
        while(!q.empty())
        {
            int u = q.front();
            q.pop();
            for(int i = head[u]; i; i = e[i].nxt) if(d[e[i].to] == -1 && e[i].f) 
            {
                d[e[i].to] = d[u] + 1;
                q.push(e[i].to);
            }
        }
        return d[sink] != -1;
    }
    int dfs(int u, int delta)
    {
        if(u == sink) return delta;
        int ret = 0;
        for(int &i = iter[u]; i && delta; i = e[i].nxt) if(d[e[i].to] == d[u] + 1 && e[i].f)
        {
            int x = dfs(e[i].to, min(delta, e[i].f));
            ret += x;
            delta -= x;
            e[i].f -= x;
            e[i ^ 1].f += x;    
        }
        return ret;
    }
    int dinic() 
    {
        int ret = 0;
        while(bfs())
        {
            for(int i = source; i <= sink; ++i) iter[i] = head[i];
            ret += dfs(source, inf);
        }
        return ret;
    }
    void link(int u, int v, int f)
    {
        e[++cnt].nxt = head[u];
        head[u] = cnt;
        e[cnt].to = v;
        e[cnt].f = f;
    }
    void insert(int u, int v, int f)
    {
        link(u, v, f);
        link(v, u, 0);
    }
    int main()
    {
        scanf("%d%d", &n, &m);
        sink = 2 * n + 1;
        for(int i = 1; i <= m; ++i) 
        {
            int u, v;
            scanf("%d%d", &u, &v);
            Map[u][v] = 1;
        }
        for(int k = 1; k <= n; ++k)
        {
            insert(source, k, 1);
            insert(k + n, sink, 1);
            for(int i = 1; i <= n; ++i)
                for(int j = 1; j <= n; ++j)
                    Map[i][j] |= (Map[i][k] & Map[k][j]);
        }
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)  if(Map[i][j])
                insert(i, j + n, 1);
        printf("%d
    ", n - dinic());                    
        return 0;
    }
    
    View Code
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  • 原文地址:https://www.cnblogs.com/19992147orz/p/8269905.html
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