Los Valley - P4997 - not Go - disjoint-set - Analog

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

The first is to change the definition of gas, making it easy to calculate that this is well understood.
Then use disjoint-set as a whole to maintain the nature of the communication block.

The final difficulty is, zi communication with the connected block color, but if the same communication block itself should not be counted, so in fact, the communication block gas should not be maintained manually, but should be disjoint-set to full power to process.

Of course, after the de-emphasis is right.

#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;

const int INF=0x3f3f3f3f;

int solve();

int main() {
#ifdef Yinku
    freopen("Yinku.in","r",stdin);
#endif // Yinku
    solve();
}

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))]--;
                    else
                        Qi++;
                }
            }

            //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;
}