次小生成树
链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=67265#overview
Problem A
ACM CONTEST AND BLACKOUT
In order to prepare the “The First National ACM School Contest”(in 20??) the major of the city decided to provide all the schools with a reliable source of power. (The major is really afraid of blackoutsJ). So, in order to do that, power station “Future” and one school (doesn’t matter which one) must be connected; in addition, some schools must be connected as well.
You may assume that a school has a reliable source of power if it’s connected directly to “Future”, or to any other school that has a reliable source of power. You are given the cost of connection between some schools. The major has decided to pick out two the cheapest connection plans – the cost of the connection is equal to the sum of the connections between the schools. Your task is to help the major – find the cost of the two cheapest connection plans.
Input
The Input starts with the number of test cases, T (1£T£15) on a line. Then T test cases follow. The first line of every test case contains two numbers, which are separated by a space, N (3£N£100) the number of schools in the city, and M the number of possible connections among them. Next M lines contain three numbers Ai, Bi, Ci , where Ci is the cost of the connection (1£Ci£300) between schools Ai and Bi. The schools are numbered with integers in the range 1 to N.
Output
For every test case print only one line of output. This line should contain two numbers separated by a single space - the cost of two the cheapest connection plans. Let S1 be the cheapest cost and S2 the next cheapest cost. It’s important, that S1=S2 if and only if there are two cheapest plans, otherwise S1£S2. You can assume that it is always possible to find the costs S1 and S2..
| Sample Input |
Sample Output |
| 2 5 8 1 3 75 3 4 51 2 4 19 3 2 95 2 5 42 5 4 31 1 2 9 3 5 66 9 14 1 2 4 1 8 8 2 8 11 3 2 8 8 9 7 8 7 1 7 9 6 9 3 2 3 4 7 3 6 4 7 6 2 4 6 14 4 5 9 5 6 10 |
110 121 37 37
|
当当子模板(写给自己看的)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
using namespace std;
int T,n,m,sum;
int s[110][110],vis[110],dis[110],used[110][110],pre[110],mx[110][110];
void prime(){
sum=0;
mst(vis,0);
mst(used,0);
mst(mx,0);
for(int i=1;i<=n;i++){
dis[i]=s[1][i];
pre[i]=1;
}
pre[1]=-1;
dis[1]=0;
vis[1]=1;
for(int i=1;i<n;i++){
int minn=maxn,flag=-1;
for(int j=1;j<=n;j++){
if(!vis[j] && dis[j]<minn){
minn=dis[j];
flag=j;
}
}
used[flag][pre[flag]]=used[pre[flag]][flag]=1;
for(int j=1;j<=n;j++){
if(vis[j])
mx[flag][j]=mx[j][flag]=max(dis[flag],mx[j][pre[flag]]);
}
vis[flag]=1;
for(int j=1;j<=n;j++){
if(!vis[j] && dis[j]>s[flag][j]){
dis[j]=s[flag][j];
pre[j]=flag;
}
}
}
for(int i=1;i<=n;i++){
sum+=dis[i];
}
}
int main(){
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(i!=j) s[i][j]=maxn;
else s[i][j]=0;
}
}
for(int i=1;i<=m;i++){
int a,b,len;
scanf("%d%d%d",&a,&b,&len);
if(s[a][b]>len)
s[a][b]=s[b][a]=len;
}
prime();
int ans=maxn;
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(i!=j && used[i][j]==0 && s[i][j]<maxn){
if(s[i][j]>=mx[i][j]){
ans=min(ans,s[i][j]-mx[i][j]);
}
}
}
}
if(ans==maxn)
printf("%d %d\n",sum,sum);
else
printf("%d %d\n",sum,sum+ans);
}
return 0;
}