HDU 2122 Ice_cream’s world III

B - Ice_cream’s world III

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

Submit Status

Description

ice_cream’s world becomes stronger and stronger; every road is built as undirected. The queen enjoys traveling around her world; the queen’s requirement is like II problem, beautifies the roads, by which there are some ways from every city to the capital. The project’s cost should be as less as better.

 

Input

Every case have two integers N and M (N<=1000, M<=10000) meaning N cities and M roads, the cities numbered 0…N-1, following N lines, each line contain three integers S, T and C, meaning S connected with T have a road will cost C.

 

Output

If Wiskey can’t satisfy the queen’s requirement, you must be output “impossible”, otherwise, print the minimum cost in this project. After every case print one blank.

 

Sample Input

      
      
      
      

       
       
       
       2 1
0 1 10

4 0 
      
      
      
      

 

Sample Output

      
      
      
      

       
       
       
       10

impossible 
      
      
      
      

 

最小生成树基础题目。

Prim算法。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define INF 0x3f3f3f3f
#define N 1005
using namespace std;
int graph[N][N];
int dis[N],vis[N];
int main()
{
        int n,m;
        while(scanf("%d%d",&n,&m)>0)
        {
                memset(vis,0,sizeof(vis));
                for(int i = 0; i < n; i++)
                {
                        graph[i][i] = 0;
                        for(int j = 0; j < i; j++)
                                graph[i][j] = graph[j][i] = INF;
                }
                int s,t,c;
                while(m--)
                {
                        scanf("%d%d%d",&s,&t,&c);
                        graph[s][t]=graph[t][s]=min(graph[s][t],c);
                }
                vis[0]=1;
                for(int i=0;i<n;i++)
                        dis[i]=graph[0][i];
                int minn,v,sum=0;
                for(int j=1;j<n;j++)
                {
                        minn=INF,v=-1;
                        v=-1;
                        for(int i=0;i<n;i++)
                                if(!vis[i])
                                        minn=minn>dis[i]?dis[v=i]:minn;
                        if(v==-1)
                                break;
                        vis[v]=1;
                        sum+=minn;
                        for(int i=0;i<n;i++)
                                if(!vis[i])
                                        dis[i] = min(graph[v][i],dis[i]);
                }
                if(v==-1)
                        printf("impossible\n");
                else
                        printf("%d\n",sum);
                printf("\n");
        }
        return 0;
}