HDU 3038

Problem 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

大致题意：

　　小屁孩玩游戏，有 n 个数，你不知道它们的值， 然后又有 m 行数，每行 a ，b ，c，表示 a 到 b 之间所有数的和为c（包含了第a个和第b个数）；

　　但是这m行数里面有些是错的，就是与前面给的条件相冲突的，要求你最后输出错了几行；

解题思路：

　　带权并查集。

　　若有 3->6 ,7->10，显然，你把7--，3--；就变成 2->10;就可以放进同一个集合了；

　　和同一个集合里的其余进行比较即可；

#include <cstdio>
using namespace std;
int sum[200005];
int f[200005];
int sf(int x){
    if(x==f[x]) return x;
    else{
        int temp=f[x];//先保存父节点再改变 
        f[x]=sf(f[x]);
        sum[x]+=sum[temp];//length: x->root = x->father + father->root
        return f[x];
    }
}
int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        for(int i=0;i<=n;i++) f[i]=i,sum[i]=0;
        int l,r,s,ans=0;
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&l,&r,&s); l--;
            int a=sf(l);//rot_l
            int b=sf(r);//rot_r
            if(a>b){
                f[a]=b;
                sum[a]=sum[r]-sum[l]-s;//a->b = r->b - l->a - l->r
            } else if(a<b){
                f[b]=a;
                sum[b]=sum[l]-sum[r]+s;
            } else {//如果是同一个集合
                if(sum[r]-sum[l]!=s)    ans++;
            }
        } printf("%d
",ans);
    } return 0;
}