HDU1301Jungle Roads

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

解题报告：很明显，此题是一个最小生成树的题目，有两种算法可以解这道题，一种是prime，另一种是克鲁斯卡尔算法，都是解最小生成树问题的算法，两种算法的区别是

首先在时间复杂度上差异很大，prime的时间复杂度是n*n，而克鲁斯卡尔算法的时间复杂度为n*logn；所以这就决定了我们在用的时候选择的问题，一般在解点较多的

最小生成树问题时，多采用克鲁斯卡尔算法，而在解点较少的问题的时候，一般选择用prime算法来解，它的优点是代码简单容易实现，而克鲁斯卡尔算法中用到了一个并查集的知识，所以要使用克鲁斯卡尔算法，必须要先熟悉并查集的知识。而克鲁斯卡尔算法的优点除了时间复杂度低一点之外，还有一个很突出的优点就是它可以记录路径，当问题要求要得出解决问题的路径时，用prime算法就显得无能为力了，而用克鲁斯卡尔算法可以很方便的记录解决问题的路径。下面具体的AC代码：

#include<iostream>
#include<stdio.h>
using namespace std;
int visit[85];
int N,sum;
struct node {
int front,rear;
int length;
}lode[85];
int find(int i,int x) {
return (i==x? x:find(x,visit[x]));
}
void popsort(int n) {
for(int i=1;i<=n;++i)
for(int j=1;j<=n-1;++j)
if(lode[j].length>lode[j+1].length) {
  struct node d=lode[j];
  lode[j]=lode[j+1];
  lode[j+1]=d;
}
}
int main() {
char jungle[5],jun[3];
int n;
while(scanf("%d",&N)&&N!=0) {
sum=0;
int z=1;
for(int i=1;i<N;++i) {
  scanf("%s%d",jun,&n);
  if(n==0)
   continue;
  for(int j=1;j<=n;++j) {
   scanf("%s%d",jungle,&lode[z].length);
   lode[z].front=jun[0]-65+1;
   lode[z++].rear=jungle[0]-65+1;
  }
}
for(int i=1;i<=N;++i)        //对visist数组进行初始化
  visit[i]=i;
popsort(z-1);
int jishu=0;
  for(int i=1;i<=z;++i)
   if(find(lode[i].front,visit[lode[i].front])!=find(lode[i].rear,visit[lode[i].rear])) {
    if(jishu>=N-1)
    break;
    jishu++;
    visit[find(lode[i].front,visit[lode[i].front])]=lode[i].rear;
    sum+=lode[i].length;
  }
  printf("%d\n",sum);
}
return 0;
}