sdnu-并查集-weeklyexam ——H - 种类并查集（选做）

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

这两个选座题真的是需要时间好好去消化去

这个与常规并查集的做法不同的是要根据这次输入和前面的输入的内容做出合并和优化

首先题要理解懂，根据一开始输入的几次区间和进行老大连接，在连老大的时候要进行和的优化sum[i]表示的是i到当前根节点的距离

pre[x] = find(f);
sum[x] += sum[f];

找到老大后如果他们老大一样，那就可以去判断这次的回答是不是错误的了

sum[l] - sum[r] != s

大值减去小值最后和输入的s进行判断

如果不想等，就得把老大连起来，并且更新sum的值，表示当小弟的老大到新老大的距离

sum[r] - sum[l] + s

也可以用向量的思想去考虑‘最后输入结束就可以统计到谎话的次数了

#include<iostream>  
#include<cstdio>  
#include<map>  
#include<cstring>  
#include<cmath>  
#include<algorithm>  
using namespace std;  
int n,m;  
int pre[200005];  
int sum[200005];  
int find(int x)  
{  
    if(x!=pre[x])  
    {  
        int f = pre[x];  
        pre[x] = find(f);  
        sum[x] += sum[f];  
    }  
    return pre[x];  
}  
int main()  
{  
    while(scanf("%d%d",&n,&m)!=EOF)  
    {  
        for(int i=0;i<=n;i++) pre[i]=i,sum[i]=0;  
        int ans=0;  
        while(m--)  
        {  
            int l,r,s;  
            scanf("%d%d%d",&l,&r,&s);  
            l--;  
            int ra = find(l),rb = find(r);  
            //cout<<"ra = "<<ra<<endl;
            //cout<<"rb = "<<rb<<endl;
            if(ra == rb)  
            {  
                if(sum[l] - sum[r] != s) ans++;  
            }  
            else  
            {  
                pre[ra] = rb;  
               // printf("pre[%d] = %d \n",ra,rb);
                sum[ra] = sum[r] - sum[l] + s;  
                //printf("sum[%d] = sum[%d] - sum[%d] + %d  =  %d\n",ra,r,l,s,sum[ra]);
            }  
        }  
        printf("%d\n",ans);  
    }  
    return 0;  
}