HDUOJ----1301 Jungle Roads

Jungle Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3968    Accepted Submission(s): 2888

Problem 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 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.

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.

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 Input

9
A 2 B 12 I 25
B 3 C 10 H 40 I 8
C 2 D 18 G 55
D 1 E 44
E 2 F 60 G 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

 

Source

Mid-Central USA 2002

最小生成树   ，解法一 普利姆算法...

做了这么多不妨做一个总结ba 。 关于普利姆算法

普利姆一般用于连续的点...（密集型的结构图）。

通过这个p大家可能会看的比较明白....

比如 第二个样列

3

A 2 B 10 C 40

B 1 C 20        a-->(b-->c  ,c)  对于这样一个图，随机从一点出发... 米前景一步变求一次最小,,,

于是 觉得用代码更有说服力。。。毕竟要说明白也很花时间,.

代码：

 // test 1 prim
 //< @ Gxjun coder>
 #include<stdio.h>
 #include<string.h>
 const int inf=0x3f3f3f3f ;
 int vis[],lowc[];
 int prim( const int cost[][], int n)
 {
     int i,j,p;
     memset(vis,,sizeof(vis));
     vis[]=;
     int minc,res=;
     for(i=; i<n ;i++)
         lowc[i]=cost[][i];
     for(i=;i<n;i++)
     {
         minc=inf;
         p=-;
         for( j=; j<n; j++)
         {
             if(==vis[j] && minc>lowc[j])
             {
                 minc = lowc[j];
                 p=j;
             }
         }
             if(minc==inf) return -;
             res+=minc;
             vis[p]=;
             for(j= ; j<n ;j++)
             {
                 if(vis[j]== && lowc[j]>cost[p][j])
                     lowc[j]=cost[p][j];
             }

     }
     return res;
 }
 int sta[][];
 int main()
 {
     int n,m,i,j;
     char a[],b[];
     int val;
     while(scanf("%d",&n),n)
     {
         for(i= ;i<; i++)
         {
             for( j=; j< ;j++)
             {
                 sta[i][j]=inf;
             }
         }
         for(i=; i<n; i++)
         {

             scanf("%s %d",a,&m);
             int fo=a[]-'A';
             while(m--)
             {
                 scanf("%s %d",b, &val);
                 int tot=b[]-'A';
                 sta[fo][tot]=sta[tot][fo]=val;
             }
         }

         printf("%d\n",prim(sta,n));
     }
     return ;
 }

方法二： 克鲁斯卡尔算法（krusal）

运用并查集进行联通分量的查找，逐步口大连通分量.....

详细代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
typedef struct Side
{
    int vs,ve;       //<vs--->表示开始的位置，ve-->代表结束的位置>
    int len ;        // <边的长度>
}side;
side sta[];       //最多50个节点
//<用到并查集来A>
 int father[];
/*初始化*/
 void mset()
 {
     for(int i=; i< ;i++)
         father[i]=i;
 }
 //<查找>
 int setfind(int x)
{
    if(x!=father[x])
        father[x]=setfind(father[x]);
    return father[x];
}
// <合并>即克鲁斯卡尔算法
 void krusal( int sx ,int sy)
 {
     int x=setfind(sx);
     int y=setfind(sy);
     if(x==y) return ;
     if(x<y)
         father[y]=x;
     else
         father[x]=y;

 }
 int cmp(const void* a ,const void* b )
 {
     return (*(side *)a).len-(*(side*)b).len;
 }

int main( void )
{
int n,m,tol,val,i,ans;
char a[],b[];
while(scanf("%d",&n),n)
{
    ans=tol=;
    for(i=; i<n ;i++)
    {
        scanf("%s%d",a,&m);
        while(m--)
        {
            scanf("%s%d",b, &val);
            sta[tol].vs=a[]-;
            sta[tol].ve=b[]-;
            sta[tol].len=val;
            tol++;
        }
    }
    mset();
    qsort(sta,tol ,sizeof(sta[]),cmp);
    for(i= ; i<tol ;i++)
    {
        int tem1=setfind(sta[i].vs);
        int tem2=setfind( sta[i].ve ) ;
        if(tem1 != tem2)  //如果两个两桶分量不在一起，那就链接起来
        {
            krusal(tem1,tem2);
            ans+=sta[i].len;
        }
    }
    printf("%d\n",ans);
}
    return ;
}