hdu 3038 How Many Answers Are Wrong (带权并查集)

题目

Input

Line 1: Two integers, N and M (1 <= N <= 200000, 1 <= M <= 40000). Means TT wrote N integers and FF asked her M questions.
Line 2..M+1: Line i+1 contains three integer: Ai, Bi and Si. Means TT answered FF that the sum from Ai to Bi is Si. It's guaranteed that 0 < Ai <= Bi <= N.
You can assume that any sum of subsequence is fit in 32-bit integer.

题意：n个数，m条信息，每条信息是 从a到b的和，为s，判断有多少条信息错误，

如果发现一条信息错误，就去掉这条信息，然后再往下看。 注意这些值可能有负的。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 200000+10;
int bin[maxn], dist[maxn], ans;

int find(int x)
{
    if(x==bin[x]) return x;
    int tmp = bin[x];
    bin[x] = find(bin[x]);
    dist[x] += dist[tmp];
    return bin[x];
}
void merge(int x, int y, int s)
{
    int fx = find(x);
    int fy = find(y);
    if(fx == fy)
    {
        if(dist[y] != dist[x] + s)
            ans++;
    }
    else
    {
        bin[fy] = fx;
        dist[fy] = dist[x] - dist[y] + s;
    }
}
int main()
{
    int n, m, i, a, b, s;
    while(~scanf("%d%d", &n, &m))
    {
        ans = 0;
        for(i = 1; i <= n; i++)
        {
            bin[i] = i;
            dist[i] = 0;
        }
        while(m--)
        {
            scanf("%d%d%d", &a, &b, &s);
            a--;  //注意
            merge(a, b, s);
        }
        printf("%d
", ans);
    }
    return 0;
}