這道題是求最小生成樹的總權值，以及生成樹任意兩點之間距離的期望。求最小生成樹就直接用Kruskal算法生成就好了，求期望一開始用的是map發現會爆內存，後來改了之後還是超時了，方法不對，正解其實是用dfs來更新每個節點的兒子個數，然後由於數據範圍比較大，一定要記得開long long還有double，下面是原題和代碼：
Abandoned country
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2735 Accepted Submission(s): 672
Problem Description
An abandoned country has n(n≤100000) villages
which are numbered from 1 to n.
Since abandoned for a long time, the roads need to be re-built. There are m(m≤1000000) roads
to be re-built, the length of each road is wi(wi≤1000000).
Guaranteed that any two wi are
different. The roads made all the villages connected directly or indirectly before destroyed. Every road will cost the same value of its length to rebuild. The king wants to use the minimum cost to make all the villages connected with each other directly or
indirectly. After the roads are re-built, the king asks a men as messenger. The king will select any two different points as starting point or the destination with the same probability. Now the king asks you to tell him the minimum cost and the minimum expectations
length the messenger will walk.
Input
The first line contains an integer T(T≤10) which
indicates the number of test cases.
For each test case, the first line contains two integers n,m indicate the number of villages and the number of roads to be re-built. Next m lines, each line have three number i,j,wi, the length of a road connecting the village i and the village j is wi.
For each test case, the first line contains two integers n,m indicate the number of villages and the number of roads to be re-built. Next m lines, each line have three number i,j,wi, the length of a road connecting the village i and the village j is wi.
Output
output the minimum cost and minimum Expectations with two decimal places. They separated by a space.
Sample Input
1
4 6
1 2 1
2 3 2
3 4 3
4 1 4
1 3 5
2 4 6
Sample Output
6 3.33
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<algorithm>
#include<queue>
#include<stack>//最小生成樹、並查集、dfs
using namespace std;
typedef long long ll;
typedef struct Edge{
int fro, to;
int w;
}Edge;
Edge e[1000010];
vector< pair<int, int> >p[100010];//只存了最小生成樹數據結構的表
int pre[100010];//記錄每個節點的前驅
int son[100010];//記錄了每個節點的兒子個數
int wt[100010];//記錄了每個子節點點所擁有的邊的邊權
int vis[100010];
bool cmp(Edge a, Edge b)
{
return a.w < b.w;
}
int find(int i)
{
if(pre[i] == i)
return i;
else
return pre[i] = find(pre[i]);
}
void join(int x, int y)
{
int fx = find(x), fy = find(y);
if(fx != fy)
pre[fx] = fy;//這裏有問題啊。。。
}
int init(int k)//更新每個節點的兒子數
{
vis[k] = 1;
int ans = 0;
for(int i = 0; i < p[k].size(); i++)
{
if(!vis[p[k][i].first])
{
wt[p[k][i].first] = p[k][i].second;
ans += init(p[k][i].first);
}
}
return son[k] = (ans+1);
}
int main()
{
int t, num = 0;
//freopen("1001.in", "r", stdin);記得刪掉文件重定向啊！！！
//freopen("1001.txt","w", stdout);
scanf("%d", &t);
while(t--)
{
int n, m;
ll maxa = 0;
ll exp = 0;
scanf("%d%d", &n, &m);
memset(son, 0, sizeof(son));
memset(vis, 0, sizeof(vis));
memset(pre, 0, sizeof(pre));
memset(wt, 0, sizeof(wt));
for(int i= 1; i <= n; i++)
p[i].clear();
for(int i = 1; i <= m; i++)
scanf("%d%d%d", &e[i].fro, &e[i].to, &e[i].w);
if(n == 1)//特殊情況要判定
{
printf("0 0\n");
continue;
}
sort(e+1, e+m+1, cmp);
for(int i = 1; i <= n; i++)
pre[i] = i;
int rst = n, ans = 0;
for(int i = 1; i <= m && rst > 1; i++)
{
int x = e[i].fro, y = e[i].to;
int fx = find(x), fy = find(y);
if(fx == fy)
continue;
else
{
join(x, y);
rst--;
maxa += e[i].w;
p[x].push_back(make_pair(y, e[i].w));//只存節點的數據結構，只能存最小生成樹的結構，不要搞錯了！！
p[y].push_back(make_pair(x, e[i].w));
}
}
int t = find(1);//尋找整個生成樹的根
init(t);//給最小生成樹建樹，並更新每個節點的子節點個數
for(int i = 1; i <= n; i++)//統計總的邊權
{
//printf("son[%d]:%d %d\n", i, son[i], wt[i]);
if(i != t)
exp += ((ll)son[i]*(n-son[i])*wt[i]);//這裏要用ll不然會爆
}
double down = (double)n*(n-1)/2;
printf("%I64d %.2f\n", maxa, exp/(double)down);
//num++;
}
return 0;
}