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洛谷P2764 最小路径覆盖问题(二分图)

2018-07-25 13:00:52来源：博客园阅读 ()

题意

给出一张有向无环图，求出用最少的路径覆盖整张图，要求路径在定点处不相交

输出方案

Sol

定理：路径覆盖 = 定点数 - 二分图最大匹配数

直接上匈牙利

输出方案的话就不断的从一个点跳匹配边

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    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;
vector<int> v[MAXN];
int link[MAXN], vis[MAXN], cnt = 1;
bool Arg(int x) {
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i];
        if(vis[to] == cnt) continue; vis[to] = cnt;
        if(!link[to] || Arg(link[to]))
            {link[to] = x; link[x] = to; return 1;}
    }
    return 0;
}
int Hunary() {
    int ans = 0;
    for(int i = 1; i <= N; i++, cnt++)
        if(Arg(i))
            ans++;
    return ans;
}
int main() {
    N = read(); M = read();
    for(int i = 1; i <= M; i++) {
        int x = read(), y = read();
        v[x].push_back(y + N);
    }
    int ans = N - Hunary();
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i <= N; i++) {
        int x = i + N;
        if(vis[i]) continue;
        do 
            printf("%d ", x = x - N);
        while(vis[x] = 1, x = link[x]);
        puts("");
    }
    printf("%d", ans);
    return 0;
}