BZOJ2437 [Noi2011]兔兔与蛋蛋 【博弈论 + 二分图匹配】

题目链接

BZOJ2437

题解

和JSOI2014很像

只不过这题动态删点

如果我们把空位置看做\(X\)的话，就会发现我们走的路径是一个\(OX\)交错的路径

然后将图二分染色，当前点必胜，当且仅当当前点必须作为最大匹配的匹配点

移动相当于删点，删点的话只需打个标记即可

判断当前点是不是必选，只需尝试更改当前点匹配点的匹配点即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 1605,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int h[maxn],ne = 1;
struct EDGE{int to,nxt;}ed[maxm];
inline void build(int u,int v){
	ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;
	ed[++ne] = (EDGE){u,h[v]}; h[v] = ne;
}
char s[45][45];
int X[4] = {0,0,-1,1},Y[4] = {-1,1,0,0};
int lk[maxn],vis[maxn],N,M,sx,sy,n;
int id[45][45],px[maxn],py[maxn],c[maxn],Out[maxn];
int win[maxn],ans[maxn],ansi;
int bd[maxn][maxn];
void dfs(int x,int y,int co){
	id[x][y] = ++n; px[n] = x; py[n] = y; c[n] = co;
	int nx,ny;
	for (int k = 0; k < 4; k++){
		nx = x + X[k];
		ny = y + Y[k];
		if (nx < 1 || ny < 1 || nx > N || ny > M || s[nx][ny] == s[x][y])
			continue;
		if (id[nx][ny]){
			if (!bd[id[nx][ny]][id[x][y]]) build(id[x][y],id[nx][ny]);
			bd[id[nx][ny]][id[x][y]] = bd[id[x][y]][id[nx][ny]] = true;
		}
		else {
			dfs(nx,ny,co ^ 1);
			if (!bd[id[nx][ny]][id[x][y]]) build(id[x][y],id[nx][ny]);
			bd[id[nx][ny]][id[x][y]] = bd[id[x][y]][id[nx][ny]] = true;
		}
	}
}
bool find(int u){
	Redge(u) if (!vis[to = ed[k].to] && !Out[to]){
		vis[to] = true;
		if (!lk[to] || find(lk[to])){
			lk[to] = u; lk[u] = to;
			return true;
		}
	}
	return false;
}
int main(){
	N = read(); M = read();
	REP(i,N) scanf("%s",s[i] + 1);
	REP(i,N) REP(j,M) if (s[i][j] == '.'){sx = i,sy = j,s[i][j] = 'X';break;}
	dfs(sx,sy,0);
	int K = read() << 1,x,y,nx = sx,ny = sy,u;
	REP(i,n) if (!c[i]) cls(vis),find(i);
	REP(i,K){
		cls(vis);
		x = read(); y = read();
		Out[u = id[nx][ny]] = true;
		if (lk[u]){
			win[i] = !find(lk[u]);
			if (win[i]) lk[lk[u]] = 0;
		}
		else win[i] = 0;
		lk[u] = 0;
		nx = x; ny = y;
	}
	win[K] = true;
	for (int i = 1; i <= K; i += 2)
		if (win[i] && win[i + 1]) ans[++ansi] = (i + 1) >> 1;
	printf("%d\n",ansi);
	REP(i,ansi) printf("%d\n",ans[i]);
	return 0;
}