【SPOJ-DYNACON1】Dynamic Tree Connectivity【LCT/并查集】

来源：互联网发布：江西省软件行业办编辑：程序博客网时间：2024/06/09 13:47

题意：

和BZOJ2049一样。

换了各种LCT姿势，一直TLE。

后来发现可以用并查集...

参考了Seter聚聚的blog：http://seter.is-programmer.com/posts/29428.html

BZOJ2049：http://blog.csdn.net/braketbn/article/details/50732912

#include <cstdio>#include <algorithm>using namespace std;const int maxn = 100005;int n, m, pre[maxn];inline int iread() {int f = 1, x = 0; char ch = getchar();for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';return f * x;}inline void makeroot(int x) {for(int y = 0, fa = pre[x]; x; fa = pre[x = fa]) {pre[x] = y;y = x;}}inline bool check(int a, int b) {for(; a != b && b; b = pre[b]);return a == b;}int main() {n = iread(); m = iread();while(m--) {char ch = getchar(); for(; ch != 'a' && ch != 'r' && ch != 'c'; ch = getchar());int a = iread(), b = iread();makeroot(a);if(ch == 'a') pre[a] = b;else if(ch == 'r') pre[b] = 0;else printf(check(a, b) ? "YES\n" : "NO\n");}return 0;}

0 0