HDU 3371 Connect the Cities-优快云博客

本文讨论了使用最小生成树算法解决由于海平面上升导致的城市连接问题。通过输入参数包括幸存城市的数量、可选连接道路的数量以及已连接城市的数量，输出所需最小费用以重建城市间的连接。实例演示了算法应用。

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4014 Accepted Submission(s): 1193

Problem Description

In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.

Input

The first line contains the number of test cases.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.

Output

For each case, output the least money you need to take, if it’s impossible, just output -1.

Sample Input

   1 6 4 3 1 4 2 2 6 1 2 3 5 3 4 33 2 1 2 2 1 3 3 4 5 6
  

Sample Output

1

Author

dandelion

Source

HDOJ Monthly Contest – 2010.04.04

Recommend

lcy

//最小生成树、就是把已经连接的边权值设为0

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <queue>
#define N 503
using namespace std;
int map[N][N];
bool b[N];
int n;
void prim()
{
   memset(b,0,sizeof(b));
   b[1]=1;
   int i,j,t=n,min,sum=0;
    while(--t)
    {    min=1000;
        for(i=2;i<=n;i++)
         if(!b[i]&&map[1][i]<min)
           min=map[1][i],j=i;
        if(min==1000) break;
        b[j]=1;
        sum+=min;
        for(i=2;i<=n;i++)
          if(!b[i]&&map[1][i]>map[j][i])
           map[1][i]=map[j][i];
    }
    if(min==1000)
     printf("-1\n");
    else
     printf("%d\n",sum);
}
int main()
{
    int m,l,k,t,T,i,j;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&l);
        for(i=1;i<=n;i++)
          for(j=1;j<=n;j++)
            map[i][j]=1000;
        while(m--)
        {
            scanf("%d%d%d",&i,&j,&k);
            map[i][j]=map[i][j]>k?k:map[i][j];
            map[j][i]=map[i][j];
        }
       while(l--)
       {
           scanf("%d%d",&t,&i);
           while(--t)
           {
               scanf("%d",&j);
               map[i][j]=map[j][i]=0;
               i=j;
           }
       }
       prim();
    }
    return 0;
}