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

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Description

TT and FF are ... friends. Uh... very very good friends -________-b

FF is a bad boy, he is always wooing TT to play the following game with him. This is a very humdrum game. To begin with, TT should write down a sequence of integers-_-!!(bored).

Then, FF can choose a continuous subsequence from it(for example the subsequence from the third to the fifth integer inclusively). After that, FF will ask TT what the sum of the subsequence he chose is. The next, TT will answer FF's question. Then, FF can redo this process. In the end, FF must work out the entire sequence of integers.

Boring~~Boring~~a very very boring game!!! TT doesn't want to play with FF at all. To punish FF, she often tells FF the wrong answers on purpose.

The bad boy is not a fool man. FF detects some answers are incompatible. Of course, these contradictions make it difficult to calculate the sequence.

However, TT is a nice and lovely girl. She doesn't have the heart to be hard on FF. To save time, she guarantees that the answers are all right if there is no logical mistakes indeed.

What's more, if FF finds an answer to be wrong, he will ignore it when judging next answers.

But there will be so many questions that poor FF can't make sure whether the current answer is right or wrong in a moment. So he decides to write a program to help him with this matter. The program will receive a series of questions from FF together with the answers FF has received from TT. The aim of this program is to find how many answers are wrong. Only by ignoring the wrong answers can FF work out the entire sequence of integers. Poor FF has no time to do this job. And now he is asking for your help~(Why asking trouble for himself~~Bad boy)

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.

Output

A single line with a integer denotes how many answers are wrong.

Sample Input

10 5
1 10 100
7 10 28
1 3 32
4 6 41
6 6 1

Sample Output

1

思路

 题意：两个小孩（女孩FF，男孩TT）在玩游戏（游戏是给出区间[a, b],算出这个区间的总和是多少），但是FF觉得游戏很无聊，就故意说错了一些值，但是TT很聪明，根据FF的回答可以推论出有的答案是相悖的。现在让你求根据已有答案，可以推论出几条是不正确的。

思路：sum[i]表示 i - 根节点的和，求[a,b]区间和用 sum[b]-sum[a-1],为表示方便，直接让 a--;

1.若 a 和 b 同属一个集合，那么需判断[a, b]区间和是否为s。

if(sum[b] - sum[a] != s) ans++;

2.若 a 和 b 不同属一个集合，把b的父亲接在a的父亲上。

sum[pb] = -sum[b]+s+sum[a];

 

#include<stdio.h>
#include<string.h>
const int maxn = 200005;
int fa[maxn],sum[maxn];

int find(int x)
{
    if (fa[x] != x)
    {
        int tmp = fa[x];
        fa[x] = find(fa[x]);
        sum[x] += sum[tmp];
    }
    return fa[x];
}

int main()
{
    int N,M;
    while (~scanf("%d%d",&N,&M))
    {
    	int res = 0;
        for (int i = 1; i <= N; i++)	fa[i] = i,sum[i] = 0;
        while (M--)
        {
            int a,b,v;
            scanf("%d%d%d",&a,&b,&v);
            a--;
            int pa = find(a),pb = find(b);
            if (pa == pb)
            {
                if (sum[b] - sum[a] != v)	res++;
            }
            else
            {
                fa[pb] = pa;
                sum[pb] = -sum[b] + v + sum[a];
            }
        }
        printf("%d\n",res);
    }
    return 0;
}