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[codevs1022]覆盖

[codevs1022]覆盖

试题描述

有一个N×M的单位方格中,其中有些方格是水塘,其他方格是陆地。如果要用1×2的矩阵区覆盖(覆盖过程不容许有任何部分重叠)这个陆地,那么最多可以覆盖多少陆地面积。

 

技术分享

输入

输入文件的第一行是两个整数NM  (1<=NM<=100),第二行为一个整数K( K<=50),接下来的K行,每行两个整数X,Y表示K个水塘的行列位置。(1<=X<=N1<=Y<=M)。

输出

输出所覆盖的最大面积块(1×2面积算一块)。

输入示例

4 4
6
1 1
1 4
2 2
4 1
4 2
4 4

输出示例

4

数据规模及约定

见“输入

题解

黑白染色后,挖去那几个被排除的点跑二分图匹配。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
	return x * f;
}

#define maxn 10010
#define maxm 160010
#define oo 2147483647
struct Edge {
	int from, to, flow;
	Edge() {}
	Edge(int _1, int _2, int _3): from(_1), to(_2), flow(_3) {}
} ;
struct Dinic {
	int n, m, s, t, head[maxn], next[maxm];
	Edge es[maxm];
	int hd, tl, Q[maxn], vis[maxn];
	int cur[maxn];
	void init(int nn) {
		n = nn; m = 0;
		memset(head, -1, sizeof(head));
		return ;
	}
	void AddEdge(int a, int b, int c) {
		es[m] = Edge(a, b, c); next[m] = head[a]; head[a] = m++;
		return ;
	}
	bool BFS() {
		memset(vis, 0, sizeof(vis)); vis[s] = 1;
		hd = tl = 0; Q[++tl] = s;
		while(hd < tl) {
			int u = Q[++hd];
			for(int i = head[u]; i != -1; i = next[i]) {
				Edge& e = es[i];
				if(!vis[e.to] && e.flow) {
					vis[e.to] = vis[u] + 1;
					Q[++tl] = e.to;
				}
			}
		}
		return vis[t] > 1;
	}
	int DFS(int u, int a) {
		if(u == t || !a) return a;
		int flow = 0, f;
		for(int& i = cur[u]; i != -1; i = next[i]) {
			Edge& e = es[i];
			if(vis[e.to] == vis[u] + 1 && (f = DFS(e.to, min(a, e.flow)))) {
				flow += f; a -= f;
				e.flow -= f; es[i^1].flow += f;
				if(!a) return flow;
			}
		}
		return flow;
	}
	int MaxFlow(int ss, int tt) {
		s = ss; t = tt;
		int flow = 0;
		while(BFS()) {
			for(int i = 1; i <= n; i++) cur[i] = head[i];
			flow += DFS(s, oo);
		}
		return flow;
	}
} sol;

#define maxs 110
bool Map[maxs][maxs];

int main() {
	int n = read(), m = read(), k = read();
	sol.init(n * m + 2); int s = n * m + 1, t = s + 1;
	for(int i = 1; i <= k; i++) {
		int a = read() - 1, b = read() - 1;
		Map[a][b] = 1;
	}
	
	for(int i = 0; i < n; i++)
		for(int j = 0; j < m - 1; j++) if(!Map[i][j] && !Map[i][j+1]) {
			int id = i * m + j + 1, rid = i * m + j + 2;
			sol.AddEdge(id, rid, 1); sol.AddEdge(rid, id, 1);
		}
	for(int i = 0; i < n - 1; i++)
		for(int j = 0; j < m; j++) if(!Map[i][j] && !Map[i+1][j]) {
			int id = i * m + j + 1, did = (i+1) * m + j + 1;
			sol.AddEdge(id, did, 1); sol.AddEdge(did, id, 1);
		}
	for(int i = 0; i < n; i++)
		for(int j = 0; j < m; j++) if(!Map[i][j]) {
			int id = i * m + j + 1;
			if((i & 1) ^ (j & 1)) sol.AddEdge(s, id, 1), sol.AddEdge(id, s, 0);
			else sol.AddEdge(id, t, 1), sol.AddEdge(t, id, 0);
		}
	
	printf("%d\n", sol.MaxFlow(s, t));
	
	return 0;
}

 

[codevs1022]覆盖