HDU 3038 How Many Answers Are Wrong(带权并查集)

HDU 3038 How Many Answers Are Wrong

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) 

InputLine 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. 
OutputA 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

题目大意：1到N是区间范围，M是有多少次询问，询问的三个数字：1 10 100 表示1到10之间的距离（？）是100，大概是这样吧：），如果给出的区间和距离与已知的有矛盾，就记录下来，问一共有多少次与前文有悖的询问。

pre：记录父节点  cnt:记录到根节点的距离

find函数：

int find(int x)             //此时find不单有查找任务，还有更新距离任务   
{  
      
    if (x==pre[x])  
        return x; 
		 
    int t=pre[x];  
    pre[x]=find(pre[x]);  
    cnt[x]+=cnt[t];       //记录到根节点的距离，一定要有一个思想，根节点是一个区间的一个端点而不是一个区间，输入的区间被合并成了两个点   
    return pre[x];         
}  

join函数：

a→x b→y 变成  a→x←y←b    s是a到b的距离

void join(int a,int b,int s)  
{  
    int x=find(a);  
    int y=find(b);  
    if (x==y)  
    {  
        if (cnt[a]+s!=cnt[b])              
            ans++;  
        return ;  
    }  
    else  
    {  
        pre[y]=x;  
        cnt[y]=cnt[a]+s-cnt[b];          //y到x的距离等于a到x的距离+b到a的距离-b到y的距离   自己动手画画图比较好理解
    }  
}  

完整代码：

#include<iostream>
using namespace std;

const int MAXN=200005;
int cnt[MAXN];
int pre [MAXN];
int ans;

void init(){

for(int i=0;i<=MAXN;i++)//n表示输入的结点的个数
{
    pre[i]=i;//将每一个结点的前导点设置为自己
    cnt[i]=0;
}

}

int find(int x)             //此时find不单有查找任务，还有更新距离任务   
{  
      
    if (x==pre[x])  
        return x; 
		 
    int t=pre[x];  
    pre[x]=find(pre[x]);  
    cnt[x]+=cnt[t];       //记录到根节点的距离，一定要有一个思想，根节点是一个区间的一个端点而不是一个区间，输入的区间被合并成了两个点   
    return pre[x];         
}  

void join(int a,int b,int s)  
{  
    int x=find(a);  
    int y=find(b);  
    if (x==y)  
    {  
        if (cnt[a]+s!=cnt[b])              
            ans++;  
        return ;  
    }  
    else  
    {  
        pre[y]=x;  
        cnt[y]=cnt[a]+s-cnt[b];          //y到x的距离等于a到x的距离+b到a的距离-b到y的距离   
    }  
}  

int main()
{
   int n,m;  
    while (cin>>n>>m)  
    {  
        int a,b,s;  
        ans=0;  
        init();  
        for (int i=1;i<=m;i++)  
        {  
            cin>>a>>b>>s;  
            join(a,b+1,s);   
        }  
        cout<<ans<<endl;   
    }  
}

BTW，在初始化的时候

const int MAXN=200005;
int cnt[MAXN];
int ans;
int pre [MAXN];

这样写不知道为什么会WA，这四条语句其他的顺序都能AC，唯独这种顺序为WA？？？

找了一晚上的错误啊啊啊啊啊啊啊啊啊啊啊啊

心态爆炸，不做题了，开始补番