hdu 4775 Infinite Go（并查集 模拟）

hdu 4775 Infinite Go

模拟围棋中棋子的气的计算和提子的判定

由于坐标数据范围较大，用map表示邻接关系；

用并查集来表示一片同色且相连的棋子中气的个数。每次新增一颗棋子，查找其四个方向的邻接棋子。若同色则并入同一集合，否则减少异色集合的气的个数（气为0则删去集合内的棋子）

#include <cstdio>#include <iostream>#include <cmath>#include <map>#include <algorithm>#include <vector>using namespace std;const int MAXN = 10010;typedef __int64 LL;#pragma comment(linker, "/STACK:1024000000,1024000000")map<pair<int,int>,int> mmp;struct _node{int qi, col;}node[MAXN];int fa[MAXN], n, cnt, sum[2], cd[MAXN];const int dir[4][2] = {0,1, 0,-1, -1,0, 1,0};int getfa(int x){if (fa[x] == x) return x; return fa[x] = getfa(fa[x]);}void init(){mmp.clear();cnt = 0;for (int i = 0; i<= n; ++i) fa[i] = i;sum[0] = sum[1] = 0;memset(cd, -1, sizeof cd);}void mdele(int col, int x, int y){cd[mmp[make_pair(x,y)]] = -1;mmp[make_pair(x,y)] = 0;for (int i = 0; i< 4; ++i){int xx = x+dir[i][0], yy = y+dir[i][1], v;if (v=mmp[make_pair(xx,yy)]){if (node[v].col == col)mdele(col,xx,yy);else{v = getfa(v);node[v].qi++;}}}}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);#endifint T;scanf("%d", &T);while (T--){scanf("%d", &n);init();for (int i = 0; i< n; ++i){int x, y, v, fu, fv, neg = 0, o;scanf("%d%d", &x, &y);mmp[make_pair(x,y)] = ++cnt;cd[cnt] = i&1;node[cnt].col = i&1;node[cnt].qi = 0;// printf("x=%d y=%d cnt=%d col=%d", x, y, cnt, node[cnt].col);for (int j = 0; j< 4; ++j){int xx = x+dir[j][0], yy = y+dir[j][1];if (xx == 0 || yy == 0){continue;}else if (v = mmp[make_pair(xx,yy)]){if (node[v].col == node[cnt].col){fu = getfa(cnt); fv = getfa(v);if (fv != fu){neg += node[fv].qi - 1;}else neg--;fa[fv] = fu;}else{int fb = getfa(v);node[fb].qi--;if (node[fb].qi == 0)mdele(node[v].col, xx, yy);}}else{neg++;}}node[cnt].qi += neg;// printf(" qi=%d\n", node[cnt].qi);if (!node[cnt].qi)mdele(node[cnt].col, x, y);}for (int i = 1; i<= cnt; ++i)sum[0]+=cd[i]==0, sum[1]+=cd[i]==1;printf("%d %d\n", sum[0], sum[1]);}    return 0;}