https://www.luogu.org/problemnew/show/P4997

首先是改变气的定义，使得容易计算，这个很好理解。
然后使用并查集，因为要维护整个连通块的性质。

最后的难点在于，落子把同颜色的连通块连接了，但假如本身就是同一个连通块则不应该计数，所以其实连通块的气不应该手动维护，而应该全权交给并查集去处理。

当然去重之后也是对的。

#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;const int INF=0x3f3f3f3f;int solve();int main() {
#ifdef Yinkufreopen("Yinku.in","r",stdin);
#endif // Yinkusolve();
}int n;
char g[605][605];inline int id(int i,int j) {return (i-1)*n+j;
}inline void aid(int f,int &i,int &j) {j=f%n;if(j==0)j=n;i=(f-j)/n+1;return;
}int fa[360005];
//并查集
int qi[360005];
//id对应的位置的气，只有并查集的根保存正确的气queue<int> Q[2];int dx[4]= {-1,1,0,0};
int dy[4]= {0,0,-1,1};char col[2]= {'X','O'};int find(int x) {int r=x;while(fa[r]!=r) {r=fa[r];}while(x!=r) {int k=fa[x];fa[x]=r;x=k;}return r;
}void merge(int x,int y) {x=find(x);y=find(y);if(x==y)return;fa[y]=x;return;
}void Exit() {puts("-1 -1");exit(0);
}inline void Show() {cout<<"G:"<<endl;for(int i=1; i<=n; i++) {cout<<g[i]+1<<endl;}cout<<endl;/*cout<<"FA:"<<endl;for(int i=1; i<=n; i++) {for(int j=1; j<=n; j++) {printf("%2d ",fa[id(i,j)]);}cout<<endl;}cout<<endl;*/cout<<"QI:"<<endl;for(int i=1; i<=n; i++) {for(int j=1; j<=n; j++) {qi[id(i,j)]=qi[find(id(i,j))];printf("%2d ",qi[id(i,j)]);}cout<<endl;}cout<<endl;
}bool RollBackOtherColor(int ox,int oy,int th,int ot) {int need=0;for(int k=0; k<4; k++) {int x=ox+dx[k];int y=oy+dy[k];if(x>=1&&x<=n&&y>=1&&y<=n&&g[x][y]==col[ot])if(qi[find(id(x,y))]<=0){need=1;break;}}if(!need)return 0;for(int k=0; k<4; k++) {int x=ox+dx[k];int y=oy+dy[k];if(x>=1&&x<=n&&y>=1&&y<=n&&g[x][y]!='.')qi[find(id(x,y))]++;}return 1;}bool RollBackThisColor(int ox,int oy,int th,int ot,int Qi) {int newQi=Qi;vector<int> visited;for(int k=0; k<4; k++) {int x=ox+dx[k];int y=oy+dy[k];/*这样同一个连通块的气被重复计算了if(x>=1&&x<=n&&y>=1&&y<=n&&g[x][y]==col[th])newQi+=qi[find(id(x,y))];*/if(x>=1&&x<=n&&y>=1&&y<=n&&g[x][y]==col[th]){int r=find(id(x,y));int s=visited.size();bool Visited=0;for(int i=0;i<s;i++){if(visited[i]==r){Visited=1;break;}}if(Visited)continue;visited.push_back(r);newQi+=qi[r];}}/*cout<<"newQi="<<newQi<<endl;cout<<"Qi="<<Qi<<endl;*/if(newQi>0) {//还有气，落子g[ox][oy]=col[th];for(int k=0; k<4; k++) {int x=ox+dx[k];int y=oy+dy[k];if(x>=1&&x<=n&&y>=1&&y<=n&&g[x][y]==col[th])merge(id(ox,oy),id(x,y));}qi[find(id(ox,oy))]=newQi;printf("%d %d\n",ox,oy);//Show();return 0;}for(int k=0; k<4; k++) {int x=ox+dx[k];int y=oy+dy[k];if(x>=1&&x<=n&&y>=1&&y<=n&&g[x][y]!='.')qi[find(id(x,y))]++;}return 1;}int Move(int th,int ot) {while(1) {//Show();//交替下while(1) {//重复下同一种棋，防爆栈if(Q[th].empty()) {Exit();}int f=Q[th].front();Q[th].pop();int x,y;aid(f,x,y);//cout<<"color="<<col[th]<<" ("<<x<<","<<y<<")"<<endl;//已经被落子，重来if(g[x][y]!='.') {//cout<<"RollBack beacuse there is already a piece"<<endl;continue;}int Qi=0;for(int k=0; k<4; k++) {int xx=x+dx[k];int yy=y+dy[k];if(xx>=1&&xx<=n&&yy>=1&&yy<=n) {if(g[xx][yy]!='.')qi[find(id(xx,yy))]--;elseQi++;}}//Show();if(RollBackOtherColor(x,y,th,ot)) {//把另一种颜色提子了，回滚重来//cout<<"RollBack by other color"<<endl;continue;}//那么下这步棋至少不会把另一种棋提子了，接下来看看是不是堵死自己if(RollBackThisColor(x,y,th,ot,Qi)) {//把自己提子了，回滚重来//cout<<"RollBack by this color"<<endl;continue;}//已经被上函数落子break;}//颜色交换swap(th,ot);}return 0;
}void Init() {//初始化并查集for(int i=1; i<=n*n; i++)fa[i]=i;//用并查集连通各部分，每个点和他右边、下边的点连接就可以了for(int i=1; i<=n; i++) {for(int j=1; j<=n; j++) {if(g[i][j]=='.')continue;if(i+1<=n&&g[i][j]==g[i+1][j])merge(id(i,j),id(i+1,j));if(j+1<=n&&g[i][j]==g[i][j+1])merge(id(i,j),id(i,j+1));}}//统计气memset(qi,0,sizeof(qi));//每个空位对各个棋子的贡献for(int i=1; i<=n; i++) {for(int j=1; j<=n; j++) {if(g[i][j]!='.')continue;//插入待使用队列Q[0].push(id(i,j));Q[1].push(id(i,j));for(int k=0; k<4; k++) {int x=i+dx[k];int y=j+dy[k];if(x>=1&&x<=n&&y>=1&&y<=n&&g[x][y]!='.')qi[find(id(x,y))]++;}}}return ;
}int solve() {scanf("%d",&n);for(int i=1; i<=n; i++) {scanf("%s",g[i]+1);}Init();Move(0,1);//先下黑棋return 0;
}