2016 Multi-University Training Contest 1 1001 hdu 5723 DFS+并查集最小生树

链接：戳这里

Abandoned country

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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.

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
 

思路：

并查集计算出最小生成树，算每条边的贡献，左右边分别有多少点

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<vector>#include <ctime>#include<queue>#include<set>#include<map>#include<stack>#include<iomanip>#include<cmath>#define mst(ss,b) memset((ss),(b),sizeof(ss))#define maxn 0x3f3f3f3f#define MAX 1000100#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef unsigned long long ull;#define INF (1ll<<60)-1using namespace std;struct node{    int u,v,w;    bool operator < (const node &a)const{        return w<a.w;    }}s[1000100];struct edge{    int v,w,next;}e[2000100];int n,m,tot;int head[1000100],val[1000100],sz[1000100],fa[1000100];void init(){    for(int i=1;i<=n;i++) fa[i]=i;    mst(sz,0);    mst(head,-1);    mst(val,0);    tot=0;}int Find(int x){    if(x!=fa[x]) return fa[x]=Find(fa[x]);    return fa[x];}void Add(int u,int v,int w){    e[tot].v=v;    e[tot].next=head[u];    e[tot].w=w;    head[u]=tot++;}void DFS(int u,int Fa){    sz[u]=1;    for(int i=head[u];i!=-1;i=e[i].next){        int v=e[i].v;        if(v==Fa) continue;        val[v]=e[i].w;        DFS(v,u);        sz[u]+=sz[v];    }}int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d%d",&n,&m);        init();        for(int i=0;i<m;i++) scanf("%d%d%d",&s[i].u,&s[i].v,&s[i].w);        sort(s,s+m);        ll an=0;        for(int i=0;i<m;i++){            int X=Find(s[i].u);            int Y=Find(s[i].v);            if(X!=Y){                fa[X]=Y;                an+=(ll)s[i].w;                Add(s[i].u,s[i].v,s[i].w);                Add(s[i].v,s[i].u,s[i].w);            }        }        DFS(1,0);        ll sum=0;        for(int i=2;i<=n;i++) {            sum+=(ll)val[i]*sz[i]*(n-sz[i]);        }        if(n==1) printf("0 0.00\n");        else printf("%I64d %.2f\n",an,sum*1.0/((ll)n*(n-1)/2));    }    return 0;}