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 = , inf = 1e9;
int rd()
{
    int x = , f = ; char c = getchar();
    while(c < '' || c > '') { if(c == '-') f = -; c = getchar(); }
    while(c >= '' && c <= '') { x = x *  + c - ''; c = getchar(); }
    return x * f;
}
int n, m, k, D, source, sink, cnt = , 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, -, sizeof(d));
    d[source] = ;
    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] == - && e[i].f)
        {
            d[e[i].to] = d[u] + ;
            q.push(e[i].to);
        }
    }
    return d[sink] != -;
}
int dfs(int u, int delta)
{
    if(u == sink) return delta;
    int ret = ;
    for(int &i = iter[u]; i && delta; i = e[i].nxt) if(d[e[i].to] == d[u] +  && e[i].f)
    {
        int x = dfs(e[i].to, min(delta, e[i].f));
        ret += x;
        delta -= x;
        e[i].f -= x;
        e[i ^ ].f += x;
    }
    return ret;
}
int dinic()
{
    int ret = ;
    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, );
}
int main()
{
    scanf("%d%d", &n, &m);
    sink =  * n + ;
    for(int i = ; i <= m; ++i)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        Map[u][v] = ;
    }
    for(int k = ; k <= n; ++k)
    {
        insert(source, k, );
        insert(k + n, sink, );
        for(int i = ; i <= n; ++i)
            for(int j = ; j <= n; ++j)
                Map[i][j] |= (Map[i][k] & Map[k][j]);
    }
    for(int i = ; i <= n; ++i)
        for(int j = ; j <= n; ++j)  if(Map[i][j])
            insert(i, j + n, );
    printf("%d\n", n - dinic());
    return ;
}