POJ 3908 Quick answer 并查集

本题是并查集的运用，在删除操作时，除了该删除的点外其他连接关系不变，那么具体操作就将这个点变成新的点即可，用一个id数组更新id即可。/* ID: sdj22251 PROG: calfflac LANG: C++ */ #include <iostream> #include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <cmath> #include <ctime> #define MAX 100000000 #define LOCA #define PI acos(-1.0) using namespace std; int pre[50005], id[50005]; int find(int x) { if(pre[x] == x) return x; int t = find(pre[x]); pre[x] = t; return t; } void join(int x, int y) { int fx = find(x); int fy = find(y); if(fx != fy) { if(fx > fy) pre[fx] = fy; else pre[fy] = fx; } } int main() { #ifdef LOCAL freopen("ride.in","r",stdin); freopen("ride.out","w",stdout); #endif int n, i, n1, n2, x, y; char s[3]; while(scanf("%d", &n) != EOF) { n1 = 0, n2 = 0; for(i = 1; i <= n; i++) { pre[i] = i; id[i] = i; } while(scanf("%s", s) != EOF) { if(s[0] == 'e') break; if(s[0] == 'c') { scanf("%d%d", &x, &y); join(id[x], id[y]); } else if(s[0] == 'd') { scanf("%d", &x); id[x] = ++n; pre[n] = n; } else if(s[0] == 'q') { scanf("%d%d", &x, &y); if(find(id[x]) == find(id[y])) n1++; else n2++; } } printf("%d , %d\n", n1, n2); } return 0; }