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poj 2594 Treasure Exploration

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题目描述

http://poj.org/problem?id=2594

简要题意：给定一个 $n$ 个点 $m$ 条边的 $DAG$，求可相交最小路径覆盖

$n\le 500,m\le 5000$

Solution

相对于不可相交最小路径覆盖，我们现在能够选择的一条边可以是原来的多条边

所以我们拿 $floyd$ 做一个传递闭包之后再求不可相交最小路径覆盖即可

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#include <iostream>
#include <cstdio>
#include <queue>
#define maxn 1010
#define INF 1000000000
using namespace std;

int n, m, g[maxn][maxn];

struct Edge {
    int to, next, w;
} e[300000]; int c1, head[maxn];
inline void add_edge(int u, int v, int w) {
    e[c1].to = v; e[c1].w = w;
    e[c1].next = head[u]; head[u] = c1++;
}

inline void Add_edge(int u, int v, int w) {
    add_edge(u, v, w); add_edge(v, u, 0);
} 

int dep[maxn], s, t; 
bool bfs() {
    fill(dep, dep + maxn, 0); dep[s] = 1;
    queue<int> Q; Q.push(s);
    while (!Q.empty()) {
        int u = Q.front(); Q.pop();
        for (int i = head[u]; ~i; i = e[i].next) {
            int v = e[i].to, w = e[i].w; 
            if (w > 0 && !dep[v]) {
                dep[v] = dep[u] + 1;
                Q.push(v); if (v == t) return 1;
            }
        }
    }
    return 0; 
}

int dfs(int u, int _w) {
    if (!_w || u == t) return _w;
    int s = 0;
    for (int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].to, w = e[i].w;
        if (w > 0 && dep[v] == dep[u] + 1) {
            int di = dfs(v, min(_w - s, w));
            e[i].w -= di; e[i ^ 1].w += di;
            s += di; if (s == _w) break;
        }
    }
    if (s < _w) dep[u] = 0;
    return s;
}

int dinic() {
    int mf = 0;
    while (bfs()) mf += dfs(s, INF);
    return mf;
}

void floyd() {
    for (int k = 1; k <= n; ++k)
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                g[i][j] |= g[i][k] & g[k][j];
}

void init() {
    fill(head, head + maxn, -1); 
    fill(g[0], g[0] + maxn * maxn, 0);
    s = 0; t = 2 * n + 1; c1 = 0; 
}

void work() { init(); 
    for (int i = 1; i <= m; ++i) {
        int x, y, z; cin >> x >> y;
        g[x][y] = 1; 
    } floyd(); 
    for (int u = 1; u <= n; ++u) {
        Add_edge(s, u, 1); Add_edge(u + n, t, 1); 
        for (int v = 1; v <= n; ++v)
            if (g[u][v]) Add_edge(u, v + n, 1);
    } cout << n - dinic() << "\n";
} 

int main() {
    //ios::sync_with_stdio(false);
    //cin.tie(nullptr); cout.tie(nullptr);

    while (cin >> n >> m && n + m) work(); 
    return 0;
}