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

 

Sample Output

 

Source

Valentine's Day Round

 

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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∗b 个选法，爱情经验值为a∗b∗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。

 #include<iostream>

 #include<cstring>

 #include<cstdlib>

 #include<cstdio>

 #include<algorithm>

 #include<cmath>

 #include<queue>

 #include<map>

 #include<set>

 #include<stack>

 #include<string>

 #define N 150005

 #define M 10005

 //#define mod 10000007

 //#define p 10000007

 #define mod2 1000000000

 #define ll long long

 #define ull unsigned long long

 #define LL long long

 #define eps 1e-6

 //#define inf 2147483647

 #define maxi(a,b) (a)>(b)? (a) : (b)

 #define mini(a,b) (a)<(b)? (a) : (b)

 using namespace std;

 int n;

 int f[N];

 ull cou[N];

 ull suma,sumi;

 int cnt;

 typedef struct

 {

     int a;

     int b;

     ull c;

 }PP;

 PP p[N];

 bool cmp(PP x,PP y)

 {

     return x.c<y.c;

 }

 int find(int x)

 {

     int fa;

     if(x!=f[x])

     {

         fa=find(f[x]);

         f[x]=fa;

     }

     return f[x];

 }

 void merge(int x,int y)

 {

     int a,b;

     a=find(x);

     b=find(y);

     if(a==b) return;

     f[b]=a;

     cou[a]=cou[a]+cou[b];

 }

 void ini()

 {

     suma=sumi=;

     int i;

     for(i=;i<=n;i++){

         f[i]=i;

         cou[i]=;

     }

     for(i=;i<n;i++){

         scanf("%d%d%I64u",&p[i].a,&p[i].b,&p[i].c);

     }

     sort(p+,p+n,cmp);

 }

 void solve()

 {

     int i;

     int aa,bb;

     for(i=;i<n;i++){

         aa=find(p[i].a);

         bb=find(p[i].b);

         suma+=cou[aa]*cou[bb]*p[i].c;

         merge(p[i].a,p[i].b);

     }

     for(i=;i<=n;i++){

         f[i]=i;

         cou[i]=;

     }

     for(i=n-;i>=;i--){

         aa=find(p[i].a);

         bb=find(p[i].b);

         sumi+=cou[aa]*cou[bb]*p[i].c;

         merge(p[i].a,p[i].b);

     }

 }

 void out()

 {

     printf("Case #%d: %I64u\n",cnt,suma-sumi);

     cnt++;

 }

 int main()

 {

     cnt=;

     //freopen("data.in","r",stdin);

     //freopen("data.out","w",stdout);

     //scanf("%d",&T);

     //for(int ccnt=1;ccnt<=T;ccnt++)

     //while(T--)

     //scanf("%d%d",&n,&m);

     while(scanf("%d",&n)!=EOF)

     {

         ini();

         solve();

         out();

     }

     return ;

 }