POJ 1251 Jungle Roads

Jungle Roads
Time Limit: 1000 MS Memory Limit: 10000 K
Total Submissions: 17586 Accepted: 7937

Description

 

The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between ages some years ago. but the jungle overtakes roads relentlessly, so the large road network is too expensive to maintain. the councel 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 versions on maintained roads, even if the route is not as short as before. the Chief Elder wocould like to tell the councer of Elders what wocould be the smallest amount they cocould spend in aacms per month to maintain roads that wowould connect all the ages. the versions are labeled A through I in the maps above. the map on the right shows the roads that cocould 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 ining only a number n, which is the number of ages, 1 <n <27, and the ages 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 ages 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 ages. the network will never have more than 75 roads. no village will have more than 15 roads going to other ages (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 ages. caution: A brute force solution that examines every possible set of roads will not finish within the one minute time limit.

Sample Input

9
A 2 B 12 I 25
B 3 C 10 H 40 I 8
C 2 D 18G 55
D 1 E 44
E 2 F 60G 38
F 0
G 1 H 35
H 1 I 35
3
A 2 B 10 C 40
B 1 C 20
0 Sample Output

216
30 questions:

There are n vertices, and each vertex is connected to several other vertices. Minimum Spanning Tree

 

This is a bare question. You should be familiar with the Prime and krusical algorithms.

1. Prime selects a vertex into the set each time, finds the shortest path from the set to the outside of the Set, and adds the vertex corresponding to the shortest path that has not entered the set to the set.

2. Krusical sorts all paths from small to large, and then adds the shortest path to the set each time (the minimum vertex of the two endpoints of the path does not enter the set ). Use and query set here

 

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int main ()//prime{    int n,i,j;    char ch;    while(cin>>n)    {        int t,s,k,sum=0;        int map[30][30]={0};        int low[30]={0},visit[30]={0};        if (n==0) break;        for (i=0; i<n; i++)            for (j=0; j<n; j++)                map[i][j]=10000;        for (i=0; i<n-1; i++)        {            cin>>ch>>t;            while(t>0)            {                cin>>ch>>s;                map[i][ch-'A']=map[ch-'A'][i]=s;                t--;            }        }        visit[0]=1;        for (i=0; i<n; i++)            low[i]=map[0][i];        for (i=1; i<n; i++)        {            int min=10000;            for (j=0; j<n; j++)                if (!visit[j] && low[j]<min)                {                    min=low[j];                    k=j;                }            sum+=low[k];            visit[k]=1;            for (j=0; j<n; j++)                if (!visit[j] && low[j]>map[k][j])                    low[j]=map[k][j];        }        cout<<sum<<endl;    }    return 0;}