HDU

题意：n个数，m次询问，每次问区间a到b之间的和为s，问有几次冲突

思路：带权并查集的应用。[a, b]和为s，所以a-1与b就能够确定一次关系。通过计算与根的距离能够推断出询问的正确性

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 200010;

int f[MAXN],arr[MAXN];
int m,n;

int find(int x) {
	if (f[x] == x)
		return x;
	int tmp = find(f[x]);
	arr[x] += arr[f[x]];
	return f[x] = tmp;
}

int main() {
	while (scanf("%d%d", &n, &m) != EOF) {
		for (int i = 0; i <= n; i++)
			f[i] = i;
		memset(arr, 0, sizeof(arr));
		int ans = 0;
		for (int t = 0; t < m; t++) {
			int a,b,s;
			scanf("%d%d%d", &a, &b, &s);
			a--;
			int f1 = find(a);
			int f2 = find(b);
			if (f1 != f2) {
				f[f2] = f1;
				arr[f2] = arr[a]+s-arr[b];
			}
			else if (arr[b]-arr[a] != s)
				ans++;
		}
		printf("%d
", ans);
	}
	return 0;
}