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

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;
}


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