HDU 4775 Infinite Go 并查集

题目链接：Infinite Go

Infinite Go

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 857    Accepted Submission(s): 279

Problem Description

　　Go is a proverbial board game originated in China. It has been proved to be the most difficult board game in the world. “The rules of Go are so elegant, organic, and rigorously logical that if intelligent life forms exist elsewhere in the universe, they almost certainly play Go.” said Emanuel Lasker, a famous chess master.
　　A Go board consists of 19 horizontal lines and 19 vertical lines. So there are 361 cross points. At the beginning, all cross points are vacant.
　　Go is played by two players. The basic rules are:
　　1. One player owns black stones and the other owns white stones.
　　2. Players place one of his stones on any vacant cross points of the board alternately. The player owns black stones moves first.
　　3. Vertically and horizontally adjacent stones of the same color form a chain.
　　4. The number of vacant points adjacent (vertically or horizontally) to a chain is called the liberty of this chain. Once the chain has no liberty, it will be captured and removed from the board.
　　5. While a player place a new stone such that its chain immediately has no liberty, this chain will be captured at once unless this action will also capture one or more enemy’s chains. In that case, the enemy’s chains are captured, and this chain is not captured.
　　
　　In effect, Go also has many advanced and complex rules. However, we only use these basic rules mentioned above in this problem.
　　Now we are going to deal with another game which is quite similar to Go. We call it “Infinite Go”. The only difference is that the size of the board is no longer 19 times 19 -- it becomes infinite. The rows are numbered 1, 2, 3, ..., from top to down, and columns are numbered 1, 2, 3, ..., from left to right. Notice that the board has neither row 0 nor column 0, which means even though the board is infinite, it has boundaries on the top and on the left.
　　In this problem, we are solving the problem that, given the actions of two players in a set of Infinite Go, find out the number of remaining stones of each player on the final board.

 

Input

　　The input begins with a line containing an integer T (1 <= T <= 20), the number of test cases.
　　For each test case, the first line contains a single integer N (1 <= N <= 10000), the number of stones placed during this set. Then follows N lines, the i-th line contains two integer X and Y (1 <= X, Y <= 2,000,000,000), indicates that the i-th stone was put on row X and column Y (i starts from 1). The stones are given in chronological order, and it is obvious that odd-numbered stones are black and even-numbered ones are white.

 

Output

　　For each test case, output two integers Nb and Nw in one line, separated by a single space. Nb is the number of black stones left on the board, while Nw is the number of white stones left on the board.

 

Sample Input

175 54 53 53 44 43 34 6

 

Sample Output

4 2

 

Source

2013 Asia Hangzhou Regional Contest 

 

题意：给出一个从（1，1）开始的无限大的棋盘，黑子白子轮流下棋，问按照围棋的规则最后双方剩多少棋。

题目分析：新加入点时需要并查集归类判断联通，同时统计每个集合周围一圈空格的数量，如果空格的数量变为0就要将这个集合删除，删除操作可以广搜遍历一遍，同时把对手棋子的集合加上空格，注意有可能出现自己下一步把自己堵死的情况，所以每次加点最后要对自身状态判断一次。

////  main.cpp//  HDU 4775 Infinite Go////  Created by teddywang on 2017/6/30.//  Copyright © 2017年 teddywang. All rights reserved.//#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<queue>using namespace std;#define MP(x,y) make_pair(x,y)typedef pair<int,int> pir;int dx[4]={0,1,0,-1};int dy[4]={1,0,-1,0};int T;int ans1,ans2;int num;int parent[100100],sum[100100],bx[100100],by[100100];map <pir,int> mp,vis;int find(int x){    if(x==parent[x]) return x;    else    {        parent[x]=find(parent[x]);        return parent[x];    }}void del(int x,int y,int c){    int nums=0;    queue<pir> q;    vis.clear();    vis[MP(x,y)]=1;    q.push(MP(x,y));    while(!q.empty())    {        pir buf=q.front();        nums++;        int fa=mp[buf];        parent[fa]=fa;        sum[fa]=0;        mp.erase(buf);        q.pop();        for(int i=0;i<4;i++)        {            int xx=buf.first+dx[i];            int yy=buf.second+dy[i];            if(xx<=0||yy<=0||vis[MP(xx,yy)]) continue;            int fb=mp[MP(xx,yy)];            if(((fb&1)^c)==0)            {                vis[MP(xx,yy)]=1;                q.push(MP(xx,yy));            }            else            {                int fs=find(fb);                sum[fs]++;            }        }    }    if(c==0) ans2-=nums;    else ans1-=nums;}int main(){    scanf("%d",&T);    while(T--)    {        scanf("%d",&num);        ans1=num/2+(num&1),ans2=num/2;        mp.clear();        for(int i=1;i<=num;i++)        {                        int x,y;            scanf("%d%d",&x,&y);            bx[i]=x;by[i]=y;            parent[i]=i;sum[i]=0;            mp[MP(x,y)]=i;            int empty=0;            for(int j=0;j<4;j++)            {                int xx=x+dx[j],yy=y+dy[j];                if(xx<=0||yy<=0) continue;                if(mp.count(MP(xx,yy))==0) empty++;                else                {                    int fa=find(mp[MP(xx,yy)]);                    sum[fa]--;                }            }            sum[i]=empty;            for(int j=0;j<4;j++)            {                int xx=x+dx[j],yy=y+dy[j];                if(xx<=0||yy<=0) continue;                if(mp.count(MP(xx,yy))==0) continue;                if(((i&1)^(mp[MP(xx,yy)]&1))==0)                {                    int fb=find(mp[MP(xx,yy)]);                    int t=find(i);                    if(fb!=t)                    {                        parent[t]=fb;                        sum[fb]+=sum[t];                    }                }                else                {                    int fb=find(mp[MP(xx,yy)]);                    if(sum[fb]==0)                        del(xx,yy,mp[MP(xx,yy)]&1);                }            }            int t=find(i);            if(sum[t]==0) del(x,y,i&1);        }        printf("%d %d\n",ans1,ans2);            }}