hdu 3038(扩展并查集)

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

题意：给出区间[1,n],下面有m组数据，l r v区间[l,r]之和为v,每输入一组数据，判断此组条件是否与前面冲突 ,最后输出与前面冲突的数据的个数.
比如 [1 5]区间和为100 然后后面给出区间[1,2]的和为 200 那肯定就是有问题的了。

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）

aaarticlea/png;base64,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" alt="" width="210" height="178" />

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)

这样说是不是懂了向量偏移的思想呢？？

下面上代码：

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int N =;
int father[N];
int sum[N];  ///记录当前结点到根结点的距离

int _find(int x){
    if(x!=father[x]){
        int t = father[x];
        father[x] = _find(father[x]);
        sum[x]+=sum[t];
    }
    return father[x];
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=;i<=n;i++){
            father[i] = i;
            sum[i] = ;
        }
        int ans = ;
        while(m--){
            int a,b,v;
            scanf("%d%d%d",&a,&b,&v);
            a--;
            int roota = _find(a);
            int rootb = _find(b);
            if(roota==rootb){
                if(sum[a]-sum[b]!=v) ans++;  ///精华部分1
            }
            else{
                father[roota] = rootb;
                sum[roota] = -sum[a]+sum[b]+v; ///精华部分2
            }
        }
        printf("%d\n",ans);
    }
    return ;
}