[ZOJ1002] Fire Net

题目链接：

走这里

题目分析：

联想到Asteroids这道题，将行和列分别作为二分图的两边的点，把点本身作为二分图的边去处理
那么这个题的限制条件打不穿的墙怎么处理呢，发现这样的话

\[...X.. \]

其实左边那段和右边那段在横着考虑的时候是互不影响的，不妨对行和列重新编号

用一下某题解的剪枝讲解的图，样例第一个大概就是这个样子
然后对于每个空地的行编号和列编号连一条边，当空地已经被控制/放满了就是最终答案，跑最大匹配即可

代码：

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <time.h>
#include <stdlib.h>
#include <string>
#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
#include <iomanip>
#define N (800 + 10)
using namespace std;
inline int read() {
	int cnt = 0, f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
	while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + c - '0'; c = getchar();}
	return cnt * f;
}
int n, cnt = 0, res = 0, l, r, ans;
char s[N][N];
int mapp[N][N], mapp2[N][N];
bool vis[N];
int nxt[N * 100], first[N * 100], to[N * 100], tot;
int match[N];
void add(int x, int y) {nxt[++tot] = first[x], first[x] = tot, to[tot] = y;}
void build_map() {
	for (register int i = 1; i <= n; ++i)
		for (register int j = 1; j <= n; ++j) {
			if (s[i][j] == 'X') {
				mapp[i][j] = 0;
				continue;
			} else {
				if (s[i][j] == '.' && s[i][j - 1] == 'X') ++cnt;
				mapp[i][j] = cnt;
			}
		}
	l = cnt, cnt = 0;
	for (register int i = 1; i <= n; ++i)
		for (register int j = 1; j <= n; ++j) {
			if (s[j][i] == 'X') {
				mapp2[j][i] = 0;
				continue;
			} else {
				if (s[j][i] == '.' && s[j - 1][i] == 'X') ++cnt;
				mapp2[j][i] = cnt;
			}
		}
	r = cnt, cnt = 0;
	for (register int i = 1; i <= n; ++i)
		for (register int j = 1; j <= n; ++j) {
			if (!mapp[i][j]) continue;
			add(mapp[i][j], mapp2[i][j]);
		}
}

bool find (int u) {
	for (register int i = first[u]; i; i = nxt[i]) {
		int v = to[i];
		if (vis[v]) continue;
		else {
			vis[v] = true;
			if (match[v] == -1 || find(match[v])) {
				match[v] = u;
				return 1;
			}
		}
	}
	return 0;
}
int hungary() {
	ans = 0;
	for (register int i = 1; i <= l; ++i) {
		memset(vis, 0, sizeof(vis));
		ans += find(i);
	}
	return ans;
}
int main() {
//	freopen("1.in", "r", stdin);
  while (1) {
  	n = read();
  	if (!n) break;
  	memset(first, 0, sizeof (first));
  	memset(to, 0, sizeof(to));
  	memset(nxt, 0, sizeof(nxt));
  	tot = 0, l = 0, r = 0;
  	for (register int i = 0; i <= n + 1; ++i)
  		for (register int j = 0; j <= n + 1; ++j) s[i][j] = 'X';
  	for (register int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
  	for (register int i = 0; i <= n + 1; ++i) s[i][n + 1] = 'X';
  	build_map();
  	for (register int i = 1; i <= r; ++i) match[i] = -1;
  	res = hungary();
  	printf("%d\n", res);
  }
	return 0;
}