HDU 4775 Infinite Go(并查集，模拟)

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HDU 4775 Infinite Go

题目链接

题意：围棋，两人轮流走，如果有一链被围死，就会被吃掉，问下完后最后黑色和白色各剩多少棋

思路：模拟，利用一个并查集来保存链，然后并记录下周围有多少个空格，然后去模拟，注意几个点，就是删除的时候，要把空格还回去，还有边界的位置是也算被围死的

代码：

#include <stdio.h>#include <string.h>#include <queue>#include <map>using namespace std;#define MP(a,b) make_pair(a,b)const int N = 10005;const int d[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};int T, n;typedef pair<int, int> pii;map<pii, int> vi, vis;int parent[N], sum[N], x[N], y[N];int find(int x) {    if (x == parent[x]) return x;    return parent[x] = find(parent[x]);}void init() {    scanf("%d", &n);    vi.clear();    for (int i = 1; i <= n; i++) {parent[i] = i; sum[i] = 0;scanf("%d%d", &x[i], &y[i]);    }}void del(int x, int y, int who) {    queue<pii> Q;    Q.push(MP(x, y));    vis.clear();    vis[MP(x,y)] = 1;    while (!Q.empty()) {pii now = Q.front();parent[vi[now]] = vi[now];sum[vi[now]] = 0;vi.erase(now);Q.pop();for (int i = 0; i < 4; i++) {    int xx = now.first + d[i][0];    int yy = now.second + d[i][1];    if (xx <= 0 || yy <= 0 || vis[MP(xx,yy)]) continue;    int tmp = vi[MP(xx,yy)];    if ((tmp&1)^who == 0) {vis[MP(xx,yy)] = 1;Q.push(MP(xx, yy));    }    else {int pt = find(tmp);sum[pt]++;    }}    }}void solve() {    for (int i = 1; i <= n; i++) {vi[MP(x[i],y[i])] = i;int empty = 0;for (int j = 0; j < 4; j++) {    int xx = x[i] + d[j][0];    int yy = y[i] + d[j][1];    if (xx <= 0 || yy <= 0) continue;    if (vi.count(MP(xx,yy)) == 0) {empty++;continue;    }    int pv = find(vi[MP(xx,yy)]);    sum[pv]--;}sum[i] = empty;for (int j = 0; j < 4; j++) {    int xx = x[i] + d[j][0];    int yy = y[i] + d[j][1];    if (xx <= 0 || yy <= 0) continue;    if (vi.count(MP(xx,yy)) == 0) continue;    if (((vi[MP(xx,yy)]&1)^(i&1)) == 0) {int pa = find(i);int pb = find(vi[MP(xx,yy)]);if (pa != pb) {    parent[pa] = pb;    sum[pb] += sum[pa];}    }    else {int pv = find(vi[MP(xx,yy)]);if (sum[pv] == 0)    del(xx, yy, vi[MP(xx,yy)]&1);    }}int pv = find(i);if (sum[pv] == 0)    del(x[i], y[i], i&1);    }    int ansa = 0, ansb = 0;    vis.clear();    for (int i = n; i >= 1; i--) {if (vi.count(MP(x[i],y[i])) == 0 || vis[MP(x[i], y[i])]) continue;vis[MP(x[i],y[i])] = 1;if (vi[MP(x[i],y[i])]&1) ansa++;else {    ansb++;}    }    printf("%d %d\n", ansa, ansb);}int main() {    scanf("%d", &T);    while (T--) {init();solve();    }    return 0;}

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