題目：
http://acm.hdu.edu.cn/showproblem.php?pid=3038
How Many Answers Are Wrong
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11657 Accepted Submission(s): 4186
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)
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.
這題是稍微變化了點的並查集，不僅僅是用並查集記錄分組情況了，還需要記錄每組一直到右端點含有的數值個數，並不斷跟新。
出現錯誤信息的判斷就是，當輸入兩個端點時候，如果兩個端點的根節點相同，就是他兩是一組的，這時候判斷數值個數是否正確，如果在已有的左端點數值+w（這段區間加上的數值個數-權值）！=右端點的數值，那一定有錯誤了，ans++。
wa的原因：find函數直接複製，忘記需要添加更新對應value值了
附ac代碼：
#include<iostream>
using namespace std;
const int maxn = 200002;
int par[maxn]; //記錄根節點，這題就是每個點區間的最左端的值，如果根節點一樣就可以判斷是不是有錯誤
int value[maxn]; //記錄每個根節點對應的數值
int find(int x)
{
if (x != par[x])
{
int pre = par[x];
par[x] = find(par[x]); //路勁壓縮
value[x] += value[pre]; //對應的值也要改變，根據之前的根節點改變
}
return par[x];
}
int main()
{
int n, m; //n個數字，m代表問的問題的個數
while (cin >> n >> m)
{
int u, v, w;
int ans = 0;
//init
for (int i = 1; i <= n; i++)
{
par[i] = i;
value[i] = 0;
}
while (m)
{
m--;
cin >> u >> v >> w;
u = u - 1; //B比A-1大S
int ru = find(u);
int rv = find(v);
if (ru == rv&&value[u] + w != value[v])
{
ans++;
}
else if (ru != rv)
{
//更新根節點,向左更新
par[rv] = ru;
value[rv] = value[u]-value[v] + w; //更新的都是右邊的端點
}
}
cout << ans << endl;
}
return 0;
}