(并查集)How Many Answers Are Wrong--HDOJ

How Many Answers Are Wrong
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 679 Accepted Submission(s): 311

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

Source
2009 Multi-University Training Contest 13 - Host by HIT

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/*    思路是参考别人的，还是有些不太明白    当根节点不同    计算距离的时候，每个点它的根节点好像都是在它们左边，    也就是比他们小的节点              --------l-------      val[u]           val[v]    |________|________|_______|    x        u        y       v    y到x的距离就是 l + val[u] - val[v]    当根节点相同    这时候只需要判断它们之间的距离是否为l就行了，    因为在一个集合内，那么点到根节点的距离就已经确定了    进而可以求出u和v之间的距离    ------val[v]-----    --val[u]--    |________|_______|    根       u       v*/#include<iostream>#include<stdio.h>#include<algorithm>#include<string>#include<string.h>#include<math.h>using namespace std;int fa[200005];int val[200005];int fid(int x){    if(x != fa[x])    {        int t = fa[x];        fa[x] = fid(fa[x]);        val[x] += val[t];    }    return fa[x];}int main(void){ //   freopen("in.txt","r",stdin);    int n,m;    while(scanf("%d %d",&n,&m) != EOF)    {        int ans = 0;        for(int i=1; i<=n; ++i)        {            fa[i] = i;            val[i] = 0;        }        for(int i=0; i<m; ++i)        {            int u,v,l;            scanf("%d %d %d",&u,&v,&l);            u--;//可以看成 u-1到 v 的距离为l            int x = fid(u);            int y = fid(v);            if(x != y)//根节点不同，就将它们归入到一个集合            {                fa[y] = x;                val[y] = val[u] + l - val[v];//y到x的距离就是u到x的距离加上l减去v到一的距离            }            else if(val[v] - val[u] != l)/                    ans++;        }        cout << ans <<endl;    }    return 0;}