hdu3038How Many Answers Are Wrong(带权并查集）

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3038

题解转载自：https://www.cnblogs.com/liyinggang/p/5327055.html

题目大意：有n个数，你不知道具体是啥，只知道有n个，然后输入m组数据，每组包含三个整数，a,b，s，表示区间[a,b]的整数和为s，输出有错误的数据的组数。

解题思路：是个典型的带权并查集，难点就在于更新相对信息值。在查找和合并也要对相对信息值进行更新。

做法是用一个数组存储某个节点到其根节点的距离。把区间的总和看成是两个端点的相对距离。

极力推荐这位大牛的博客：http://blog.csdn.net/niushuai666/article/details/6981689

这题让我学到了很多，特别是关于向量偏移，可以直接找到根节点与子节点的关系

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TzfvMRFKhWHp3qeNxcUkFVsunUz1uSXvpBfbLoVNOo3za5noLK5NCp7mdO/C8PoDa5dOq3TD9/99MpqEgOnWp/KuF6f3rdB5XJo1NAzXQKiE6ngOh0CohOp4DodAqITqeA6HQKiE6ngOh0CohOp4DodAqITqeA6HQKiE6ngOh0CohOp4DodAqITqeA6HQKiE6ngOh0CohOp4DodAqITqeA6HQKiE6ngOh0CohOp4DodAqITqeA6HQKiE6ngOh0CohOp4DodAqITqeA6HQKiE6ngOh0CohOp4DodAqITqeA6HQKiE6ngOh0CohOp4DodAqITqeA6HQKiE6ngOh0CohOp4DodAqITqeA6HQKiE6ngOh0CohOp4Do/gcBSYLaE1UYqwAAAABJRU5ErkJggg==" alt="" width="262" height="220" />

这题我们利用一个sum[]数组保存从某点到其祖先节点距离。

1.从上图我们可以看出，当roota!=rootb时 如果将roota并入rootb，那么是不是 roota->rootb = b->rootb - b->roota

然后我们可以知道 b->roota = a->roota - a->b

所以最后可以推出 roota ->rootb = b->rootb + a->b - a->roota

而roota的根节点是rootb,所以 roota->rootb = sum[roota]

然后依次推出得到 sum[roota] = -sum[a]+sum[b]+v （这里的a要说明一下由于是区间 [a,b] ，[a,b] = [root,b]-[root,a-1],所以a要减一）

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2.如果roota==rootb 是不是 a和b的根节点已经相同了？所以我们只要验证 a->b是否与题中的长度一致了。

所以 a->b = a->root - b->root

然后得到表达式 v = sum[a]-sum[b] (一定要记住这里的sum都是相对于根节点的，sum的更新在路径压缩的时候更新了)

 #include<stdio.h>
 #include<string.h>
 #include<iostream>
 using namespace std;
 const int maxn=;
 int par[maxn],rank[maxn],ans;

 int find(int x)
 {
     if(par[x]==x)
         return x;
     int temp=find(par[x]);
     rank[x]+=rank[par[x]];
     return par[x]=temp;
 }

 void unite(int x,int y,int s)
 {
     int rootx=find(x);
     int rooty=find(y);
     if(rootx!=rooty)
     {
         par[rooty]=rootx;
         rank[rooty]=rank[x]+s-rank[y];  //用向量的思维，合并后rooy到rootx的距离就等于x到根距离加上x和y的距离减去y到rooty的距离
     }
     else
     {
         if(rank[x]+s!=rank[y])
             ans++;
     }
 }

 signed main()
 {
     int n,m;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         for(int i=;i<=n;i++)
             par[i]=i,rank[i]=;
         ans=;
         while(m--)
         {
             int a,b,s;
             scanf("%d%d%d",&a,&b,&s);
             a--;
             unite(a,b,s);
         }
         printf("%d\n",ans);
     }
     return ;
 }