POJ - 1251----Jungle Roads(prim)

The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so the large road network is too expensive to maintain. The Council of Elders must choose to stop maintaining some roads. The map above on the left shows all the roads in use now and the cost in aacms per month to maintain them. Of course there needs to be some way to get between all the villages on maintained roads, even if the route is not as short as before. The Chief Elder would like to tell the Council of Elders what would be the smallest amount they could spend in aacms per month to maintain roads that would connect all the villages. The villages are labeled A through I in the maps above. The map on the right shows the roads that could be maintained most cheaply, for 216 aacms per month. Your task is to write a program that will solve such problems.

Input
The input consists of one to 100 data sets, followed by a final line containing only 0. Each data set starts with a line containing only a number n, which is the number of villages, 1 < n < 27, and the villages are labeled with the first n letters of the alphabet, capitalized. Each data set is completed with n-1 lines that start with village labels in alphabetical order. There is no line for the last village. Each line for a village starts with the village label followed by a number, k, of roads from this village to villages with labels later in the alphabet. If k is greater than 0, the line continues with data for each of the k roads. The data for each road is the village label for the other end of the road followed by the monthly maintenance cost in aacms for the road. Maintenance costs will be positive integers less than 100. All data fields in the row are separated by single blanks. The road network will always allow travel between all the villages. The network will never have more than 75 roads. No village will have more than 15 roads going to other villages (before or after in the alphabet). In the sample input below, the first data set goes with the map above.
Output
The output is one integer per line for each data set: the minimum cost in aacms per month to maintain a road system that connect all the villages. Caution: A brute force solution that examines every possible set of roads will not finish within the one minute time limit.

Sample Input9A 2 B 12 I 25B 3 C 10 H 40 I 8C 2 D 18 G 55D 1 E 44E 2 F 60 G 38F 0G 1 H 35H 1 I 353A 2 B 10 C 40B 1 C 200Sample Output21630

题意：在相通n个岛屿的所有桥都坏了，要重修，重修每一个桥所用的时间不同，求重修使每个岛屿都间接或直接与其他岛屿相同时所用的的最短时间（只有修完一个桥后才可修下一个桥）。
思路：说白了就是求最小生成树 prim算法搞定

#include<cstdio> #include<iostream>#include<algorithm> using namespace std;const int MAX = 30;const int INF = 0x3f3f3f3f;//最小生成树prim算法 struct A{    int arcs[MAX][MAX];//权值-->花费的时间    int arcnum;        //边的数目     bool visit[MAX];   //顶点的访问情况     int vexnum;        //顶点的数目 }G;int d[MAX];//选中集与未选中集的距离 int prim(void){      int min,v,ans=0;      for(int i=0 ; i<G.vexnum ; i++){          d[i] = G.arcs[0][i];         G.visit[i] = false;//每个点未访问     }    //起始点加入选中集     G.visit[0]=true;    for(int i=1 ; i<G.vexnum ; i++){//选n-1个定点         //找出一个最短距离的点         min=INF;        for(int j=0 ; j<G.vexnum ; j++){              if(!G.visit[j] && d[j]<min){                  v=j;                  min = d[j];              }          }          //已访问         G.visit[v]=true;        ans += min;        //更新选中点集与未选中点集的距离        for(int j=0 ; j<G.vexnum ; j++){            if(!G.visit[j] && G.arcs[v][j]<d[j]){                      d[j]= G.arcs[v][j];              }        }     }      return ans;  }  int main(void){      int m,weight;      char a[2],b[2];      while(cin>>G.vexnum,G.vexnum){          //初始化图         for(int i=0 ; i<G.vexnum ; i++){            for(int j=0 ; j<G.vexnum ; j++){                G.arcs[i][j] = (i==j)?0:INF;              }        }        //赋权值         for(int i=1 ; i<G.vexnum ; i++){              cin>>a>>m;              for(int j=0 ; j<m ; j++){                  cin>>b>>weight;                 G.arcs[a[0]-'A'][b[0]-'A']=weight;                 G.arcs[b[0]-'A'][a[0]-'A']=weight;              }          }         cout<<prim()<<endl;      }      return 0;  }