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  • 「Luogu1402」酒店之王

    传送门
    Luogu

    解题思路

    网络流板子题。
    建图细节见代码,也可以参考这道差不多的题

    细节注意事项

    • 咕咕咕。

    参考代码

    #include <algorithm>
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <cctype>
    #include <cmath>
    #include <ctime>
    #include <queue>
    #define rg register
    using namespace std;
    template < class T > inline void read(T& s) {
    	s = 0; int f = 0; char c = getchar();
    	while (!isdigit(c)) f |= c == '-', c = getchar();
    	while (isdigit(c)) s = s * 10 + c - 48, c = getchar();
    	s = f ? -s : s;
    }
    
    const int _ = 502;
    const int __ = 30002;
    const int INF = 2147483647;
    
    int tot = 1, head[_], nxt[__ << 1], ver[__ << 1], cap[__ << 1];
    inline void Add_edge(int u, int v, int d)
    { nxt[++tot] = head[u], head[u] = tot, ver[tot] = v, cap[tot] = d; }
    inline void link(int u, int v, int d) { Add_edge(u, v, d), Add_edge(v, u, 0); }
    
    int n, p, q, s, t, dep[_], cur[_];
    
    inline int bfs() {
    	static queue < int > Q;
    	memset(dep, 0, sizeof (int) * (t - s + 1));
    	dep[s] = 1, Q.push(s);
    	while (!Q.empty()) {
    		int u = Q.front(); Q.pop();
    		for (rg int i = head[u]; i; i = nxt[i]) {
    			int v = ver[i];
    			if (dep[v] == 0 && cap[i] > 0)
    				dep[v] = dep[u] + 1, Q.push(v);
    		}
    	}
    	return dep[t] > 0;
    }
    
    inline int dfs(int u, int flow) {
    	if (u == t) return flow;
    	for (rg int& i = cur[u]; i; i = nxt[i]) {
    		int v = ver[i];
    		if (dep[v] == dep[u] + 1 && cap[i] > 0) {
    			int res = dfs(v, min(flow, cap[i]));
    			if (res) { cap[i] -= res, cap[i ^ 1] += res; return res; }
    		}
    	}
    }
    
    inline int Dinic() {
    	int res = 0;
    	while (bfs()) {
    		for (rg int i = s; i <= t; ++i) cur[i] = head[i];
    		while (int d = dfs(s, INF)) res += d;
    	}
    	return res;
    }
    
    int main() {
    #ifndef ONLINE_JUDGE
    	freopen("cpp.in", "r", stdin);
    	freopen("cpp.out", "w", stdout);
    #endif
    	read(n), read(p), read(q);
    	s = 0, t = p + 2 * n + q + 1;
    	int f;
    	for (rg int i = 1; i <= n; ++i)
    		for (rg int j = 1; j <= p; ++j) {
    			read(f); if (f) link(j, i + p, 1);
    		}
    	for (rg int i = 1; i <= n; ++i)
    		for (rg int j = 1; j <= q; ++j) {
    			read(f); if (f) link(i + p + n, j + p + 2 * n, 1);
    		}
    	for (rg int i = 1; i <= n; ++i) link(i + p, i + p + n, 1);
    	for (rg int i = 1; i <= p; ++i) link(s, i, 1);
    	for (rg int i = 1; i <= q; ++i) link(i + p + 2 * n, t, 1);
    	printf("%d
    ", Dinic());
    	return 0;
    }
    

    完结撒花 (qwq)

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  • 原文地址:https://www.cnblogs.com/zsbzsb/p/11827417.html
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