【luogu P1402 酒店之王】 题解

题目链接:https://www.luogu.org/problemnew/show/P1402

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e6 + 10;
const int inf = 1e9;
int n, p, q, s, t, deep[maxn], maxflow;
struct edge{
	int flow, next, to;
}e[maxn<<2];
int head[maxn], cnt = -1;
queue<int> q1;
void add(int u, int v, int w)
{
	e[++cnt].flow = w; e[cnt].next = head[u]; e[cnt].to = v; head[u] = cnt;
	e[++cnt].flow = 0; e[cnt].next = head[v]; e[cnt].to = u; head[v] = cnt;
}
bool bfs(int s, int t)
{
	memset(deep, 0x7f, sizeof(deep));
	while(!q1.empty()) q1.pop();
	q1.push(s); deep[s] = 0;
	while(!q1.empty())
	{
		int now = q1.front(); q1.pop();
		for(int i = head[now]; i != -1; i = e[i].next)
		{
			if(deep[e[i].to] > inf && e[i].flow)
			{
				deep[e[i].to] = deep[now] + 1;
				q1.push(e[i].to);
			}
		}
	}
	if(deep[t] < inf) return true;
	else return false;
}
int dfs(int now, int t, int limit)
{
	if(!limit || now == t) return limit;
	int flow = 0, f;
	for(int i = head[now]; i != -1; i = e[i].next)
	{
		if(deep[e[i].to] == deep[now] + 1 && (f = dfs(e[i].to, t, min(e[i].flow, limit))))
		{
			flow += f;
			limit -= f;
			e[i].flow -= f;
			e[i^1].flow += f;
			if(!limit) break;
		}
	}
	return flow;
}
void Dinic(int s, int t)
{
	while(bfs(s, t))
	maxflow += dfs(s, t, inf);
}
int main()
{
	memset(head, -1, sizeof(head));
	scanf("%d%d%d",&n,&p,&q);
	s = 1, t = n + n + p + q + 2;
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= p; j++)
		{
			int u;
			scanf("%d",&u);
			if(u == 1)
			{
				add(j + n + n + 1, i + 1, 1);
			}
		}
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= q; j++)
		{
			int u;
			scanf("%d",&u);
			if(u == 1)
			{
				add(i + n + 1, j + n + n + p + 1, 1);
			}
		}
	for(int i = 1; i <= n; i++)
	add(i + 1, i + n + 1, 1);
	for(int i = 1; i <= p; i++)
	add(s, i + n + n + 1, 1);
	for(int i = 1; i <= q; i++)
	add(i + n + n + p + 1, t, 1);
	Dinic(s, t);
	printf("%d",maxflow);
	return 0;
}
posted @ 2018-08-09 10:41  Misaka_Azusa  阅读(77)  评论(0编辑  收藏  举报
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