uva 10600 最小与次小生成树
Problem A
ACM CONTEST AND BLACKOUT
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 |
求最小生成树与次小生成树。
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 105
#define inf 99999999
int map[N][N],F[N][N];//F存储第i个节点到第j个节点的路径中的最大权值
int vis[N],dis[N],pre[N];
int use[N][N];//标记边Eij,0表示不存在，1表示用过，2表示存在没用过
int n,m;
int sum1,sum2;
int prime()
{
sum1=0;
memset(vis,0,sizeof(vis));
int k=1;
for(int i=1; i<=n; i++)
{
vis[i]=0;
dis[i]=map[1][i];
pre[i]=1;
}
pre[1]=-1;
vis[1]=1;
for(int i=1; i<n; i++)
{
int mmin=inf;
for(int j=1; j<=n; j++)
{
if(!vis[j]&&dis[j]<mmin)
{
mmin=dis[j];
k=j;
}
}
if(pre[k]!=-1)
{
use[pre[k]][k]=use[k][pre[k]]=1;
for(int j=1; j<=n; j++)
{
if(vis[j])
F[j][k]=max(F[j][pre[k]],map[pre[k]][k]);
}
}
vis[k]=1;
sum1+=mmin;
for(int j=1;j<=n;j++)
{
if(!vis[j]&&dis[j]>map[k][j])
{
dis[j]=map[k][j];
pre[j]=k;
}
}
}
return sum1;
}
int second()
{
sum2=inf;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
if(use[i][j]==2&&sum1+map[i][j]-F[i][j]<sum2)
{
sum2=sum1+map[i][j]-F[i][j];
}
}
return sum2;
}
int main()
{
int cas,a,b,v;
cin>>cas;
while(cas--)
{
cin>>n>>m;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
map[i][j]=inf;
F[i][j]=0;
use[i][j]=0;
}
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&a,&b,&v);
map[a][b]=map[b][a]=v;
use[a][b]=use[b][a]=2;
}
int s1=prime();
int s2=second();
printf("%d %d\n",s1,s2);
}
return 0;
}
/*
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
*/