bzoj 2437 [Noi2011]兔兔与蛋蛋 二分图博弈

题面

题目传送门

解法

不妨把问题转化成空格的移动，将一开始空格的位置变成黑色

那么，真实颜色和后来染的颜色不同的就一定不会经过

然后就变成了一个二分图博弈问题了

考虑一下什么时候先手必胜：

就是起点一定在二分图的最大匹配上的时候

否则后手必胜

先手走错的情况满足在他走之前先手必胜，走之后先手还是必胜

匈牙利算法即可

代码

#include <bits/stdc++.h>
#define N 50
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Edge {
	int next, num;
} e[N * N * N * N];
int dx[5] = {0, -1, 1, 0, 0}, dy[5] = {0, 0, 0, -1, 1};
int n, m, cnt, tot, tim;
int num[N][N], col[N][N], s[N * N], p[N * N], ban[N * N], vis[N * N], ans[N * N];
void add(int x, int y) { 
	e[++cnt] = (Edge) {e[x].next, y};
	e[x].next = cnt;
}
bool dfs(int x) {
	if (ban[x]) return false;
	for (int p = e[x].next; p; p = e[p].next) {
		int k = e[p].num;
		if (vis[k] == tim || ban[k]) continue;
		vis[k] = tim;
		if (!ans[k] || dfs(ans[k])) {
			ans[k] = x, ans[x] = k;
			return true;
		}
	}
	return false;
}
int main() {
	read(n), read(m);
	int x, y;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++) {
			char c; cin >> c;
			if (c == '.') x = i, y = j, col[x][y] = 1;
			if (c == 'X') col[i][j] = 1;
		}
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			if ((i + j) % 2 != (x + y) % 2 && col[i][j] == 1 || (i + j) % 2 == (x + y) % 2 && !col[i][j])
				col[i][j] = 2;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			if (col[i][j] != 2) num[i][j] = ++tot;
	cnt = tot;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++) {
			if (col[i][j] == 2) continue;
			for (int k = 1; k <= 4; k++) {
				int tx = i + dx[k], ty = j + dy[k];
				if (tx <= 0 || ty <= 0 || tx > n || ty > m || col[tx][ty] == 2) continue;
				add(num[i][j], num[tx][ty]);
			}
		}
	for (int i = 1; i <= tot; i++)
		if (!ans[i]) tim++, dfs(i);
	int q; read(q);
	for (int i = 1; i <= q * 2; i++) {
		ban[num[x][y]] = 1;
		if (ans[num[x][y]]) {
			int tmp = ans[num[x][y]];
			ans[tmp] = ans[num[x][y]] = 0;
			tim++; p[i] = !dfs(tmp);
		} else p[i] = 0;
		read(x), read(y);
	}
	int ret = 0;
	for (int i = 2; i <= q * 2; i += 2)
		if (p[i] && p[i - 1]) s[++ret] = i / 2;
	cout << ret << "\n";
	for (int i = 1; i <= ret; i++) cout << s[i] << "\n";
	return 0;
}