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

思路：

首先，应该只有 存在两组连续区间相同时 才能判断是否出错。

多个区间连续时，使其中一个为根节点，计算每个点到根节点权值。

如果left-1与r两者根节点相同，即可以判断是否正确，如果不同使其中一个变为根节点，计算另一点到此点权值。PS.权值计算时候是有方向的要注意。

代码：

#include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> #include<algorithm> #include<queue> using namespace std; #define eps 1e-8 #define mod 1000000007 #define ll long long #define INF 0x3f3f3f3f #define mem(a,b) memset(a,b,sizeof(a)) int f[200005],sum[200005];//f表示父亲节点下标，sum表示该节点到根节点的权值 int getf(int x){ if(f[x]==x)return x; int t=f[x]; f[x]=getf(f[x]); sum[x]+=sum[t]; //回溯时顺便得到该点到该根节点距离 return f[x]; } int main(){ int n,m,i,l,r,v,t1,t2; while(~scanf("%d%d",&n,&m)){ //初始化，从0开始因为 left 要-1 int ans=0; for(i=0;i<=n;i++){ f[i]=i; sum[i]=0; } while(m--){ scanf("%d%d%d",&l,&r,&v); l--; t1=getf(l),t2=getf(r); if(t1!=t2){ // f[t1]=t2; // sum[t1]=-sum[l]+(sum[r]-v); f[t2]=t1; sum[t2]=sum[l]-(sum[r]-v); } else if(sum[l]+v!=sum[r])ans++; } printf("%d\n",ans); } return 0; }

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