【杭电oj】3038 - How Many Answers Are Wrong（带权并查集，好题）

How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4436    Accepted Submission(s): 1700

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

首先这道题有多组数据，题目没说明，小心！

这道题很好的学习了一下数据结构，写代码的时候仿佛数据构成了一片网，反正是纠结了半天。

说明一下，输入子序列 a , b 开头和结尾，不能就按照这个区间计算，应该把左区间减一，表示b-（a-1）的值，这样才方便合并。

我觉得根据代码方便理解吧，

代码如下：

#include <cstdio>
#include <cstring>
int f[200022],sum[200022];
int find(int x)		//此题关键 
{
	if (x!=f[x])
	{
		int t;
		t=f[x];		//t为其父节点 
		f[x]=find(f[x]);		//递归找根 
		sum[x]+=sum[t];
	}
	return f[x];
}

int main()
{
	int n,m;
	int x,y,tsum;
	int ans;
	while (~scanf ("%d%d",&n,&m))
	{
		for (int i=1;i<=n;i++)		//初始化 
			f[i]=i;
		ans=0;
		memset (sum,0,sizeof(sum));		//初始化，每个节点都是独立的，其sum值为0 
		for (int i=1;i<=m;i++)
		{
			scanf ("%d %d %d",&x,&y,&tsum);
			x--;
			int fx,fy;
			fx=find(x);
			fy=find(y);
			if (fx==fy)
			{
				if (tsum!=sum[y]-sum[x])
					ans++;
			}
			else
			{
				f[fy]=fx;		//把根结点连过去
				//sum[y]=sum[x]+tsum;
				//此步骤错误，只需要计算节点到其父节点的sum值即可 
				sum[fy]=sum[x]+tsum-sum[y];
			}
		}
		printf ("%d\n",ans);
	}
	return 0;
}