hdu 5176 The Experience of Love（并查集）

The Experience of Love

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 645    Accepted Submission(s): 216

Problem Description

A girl named Gorwin and a boy named Vivin is a couple. They arrived at a country named LOVE. The country consisting of N cities and only N−1 edges (just like a tree), every edge has a value means the distance of two cities. They select two cities to live，Gorwin living in a city and Vivin living in another. First date, Gorwin go to visit Vivin, she would write down the longest edge on this path(maxValue).Second date, Vivin go to Gorwin, he would write down the shortest edge on this path(minValue),then calculate the result of maxValue subtracts minValue as the experience of love, and then reselect two cities to live and calculate new experience of love, repeat again and again.

Please help them to calculate the sum of all experience of love after they have selected all cases.

 

Input

There will be about 5 cases in the input file.
For each test case the first line is a integer N, Then follows n−1 lines, each line contains three integers a, b, and c, indicating there is a edge connects city a and city b with distance c.

[Technical Specification]
1<N<=150000,1<=a,b<=n,1<=c<=109

 

Output

For each case，the output should occupies exactly one line. The output format is Case #x: answer, here x is the data number, answer is the sum of experience of love.

 

Sample Input

3
1 2 1
2 3 2
5
1 2 2
2 3 5
2 4 7
3 5 4

 

Sample Output

Case #1: 1
Case #2: 17

Hint
huge input,fast IO method is recommended.

In the first sample:
The maxValue is 1 and minValue is 1 when they select city 1 and city 2, the experience of love is 0.
The maxValue is 2 and minValue is 2 when they select city 2 and city 3, the experience of love is 0.
The maxValue is 2 and minValue is 1 when they select city 1 and city 3, the experience of love is 1.
so the sum of all experience is 1;
 

题意：给一棵树，求任意｛两点路径上的最大边权值-最小边权值｝的总和。 解法：$sigma(maxVal[i]-minVal[i])=sigma(maxVal)-sigma(minVal)$;所以我们分别求所有两点路径上的最大值的和，还有最小值的和。再相减就可以了。求最大值的和的方法用带权并查集，把边按权值从小到大排序，一条边一条边的算，当我们算第$i$条边的时候权值为$wi$,两点是$ui,vi$,前面加入的边权值一定是小于等于当前$wi$的，假设与$ui$连通的点有$a$个，与$vi$连通的点有$b$个，那么在$a$个中选一个，在$b$个中选一个，这两个点的路径上最大值一定是$wi$,一共有$a\ast b$个选法，爱情经验值为$a\ast b\ast wi$。 求最小值的和的方法类似。 槽点： 一：这题做数据的时候突然想到的把数据范围设在 unsigned long long 范围内，要爆 long long，这样选手在wa了之后可能心态不好找不到这个槽点，当是锻炼大家的心态和出现wa时的找错能力了，把这放在pretest..很良心的。 二，并查集的时候，用是递归的需要扩栈，一般上$10w$的递归都需要，所以看见有几个FST在栈溢出的，好桑心。

注意此题要用无符号长整形= = 150000*75000*75000*1000000000会爆long long

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
#define N 200050
struct Edge
{
    unsigned long long u,v,w;
}edge[N];
int n;
unsigned long long f[N],num[N];
unsigned long long maxn,minn;
bool cmp(Edge a,Edge b)
{
    return a.w<b.w;
}
void init()
{
    for(int i=1;i<=n;i++)
    {
        f[i]=i;
        num[i]=1;
    }
}
unsigned long long finds(unsigned long long x)
{
    unsigned long long r=x;
    while(f[r]!=r)
        r=f[r];
    while(f[x]!=x)
    {
        unsigned long long m=x;
        x=f[x];
        f[m]=r;
    }
    return r;
}
void Union(int id,int flag)
{
    unsigned long long u=edge[id].u,v=edge[id].v,w=edge[id].w;
    unsigned long long x=finds(u),y=finds(v);
    if(flag) maxn+=num[x]*num[y]*w;
    else minn+=num[x]*num[y]*w;
    f[x]=y;
    num[y]+=num[x];
}
int main()
{
    int t=1;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<n;i++)
            scanf("%I64u %I64u %I64u",&edge[i].u,&edge[i].v,&edge[i].w);
            sort(edge+1,edge+n,cmp);
        init();
        maxn=0,minn=0;
        for(int i=1;i<n;i++)
            Union(i,1);
        init();
        for(int i=n-1;i>=1;i--)
            Union(i,0);
        printf("Case #%d: %I64u\n",t++,maxn-minn);
    }
    return 0;
}