HDU3038 How Many Answers Are Wrong (加权并查集)

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 51 10 1007 10 281 3 324 6 416 6 1

Sample Output

1

题意：输入n，m。接下来m行，每行输入A，B，S，（0<=A,B<=n）表示从A到B这（B-A+1）个数的和为S。问这m句话中有几句是错误的。

分析；用加权并查集（加权并查集与普通的并查集之间区别就是，增加啦一个数组表示该节点与其父节点的权值，在查找根节点和合并两个节点时，维护一下，就ok）

#include<cstring>#include<cstdio>#include<iostream>using namespace std;int n,m,f[202000],v[202000];void init(){    for(int i=0;i<=n;i++)    {        f[i]=i;        v[i]=0;    }}int findd(int x)  {    int tem=f[x];    if(x==f[x])return x;    f[x]=findd(f[x]);    v[x]=v[x]+v[tem]; // 通过递归过程，当执行到这一步时，x的根结点是其爷爷节点。那么v[x]=x[x]+v[tem];    return f[x];}int main(){    int x,y,w,fx,fy,ans;    while(scanf("%d%d",&n,&m)==2)    {        ans=0;        init();        while(m--)        {            scanf("%d%d%d",&x,&y,&w);            x--;            fx=findd(x);            fy=findd(y);         //cout<<fx<<"  *******"<<fy<<endl;            if(fx!=fy)            {                f[fx]=fy;                v[fx]=v[y]-v[x]+w;// 此时x的根结点是fx，y的根节点是fy 则v[x]+v[fx]=w+v[y];            }            else            {                if(v[x]!=(v[y]+w))ans++;            }         // for(int i=0;i<=n;i++)findd(i);           // for(int i=0;i<=n;i++)cout<<v[i]<<"  ";cout<<endl;        }        printf("%d\n",ans);    }    return 0;}