HDU 4775 Infinite Go
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; }
HDU 4775 Infinite Go(并查集,模拟)
时间: 2024-10-07 09:08:49