题目：
Jungle Roads |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 126 Accepted Submission(s): 118 |
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Problem Description
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. |
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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 |
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Sample Output 216 30 |
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Source Mid-Central USA 2002 |
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Recommend Eddy |
题目分析：
使用kruscal求最小生成树。这道题需要注意以下几点：
1）这道题给出的是字符索引，我们需要将字符索引转数字索引。如：edges[cnt].begin = start - ‘A‘ + 1;
2）另外，这道题的输入数据的格式也需要注意一下。
A 2 B 12 I 25 B 3 C 10 H 40 I 8
这组输入样例的意思是：
A与2个点相连，与点B相连的边AB的权值为12，与点I相连的权值是25
代码如下：
/*
* f.cpp
*
* Created on: 2015年3月10日
* Author: Administrator
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 28;
struct Edge{
int begin;
int end;
int weight;
}edges[maxn*maxn];
int father[maxn];
int find(int a){
if(a == father[a]){
return a;
}
return father[a] = find(father[a]);
}
int kruscal(int count){
int i;
for(i = 1 ; i < maxn ; ++i){
father[i] = i;
}
int sum = 0;
for(i = 1 ; i <= count ; ++i){
int fa = find(edges[i].begin);
int fb = find(edges[i].end);
if(fa != fb){
father[fa] = fb;
sum += edges[i].weight;
}
}
return sum;
}
bool cmp(Edge a,Edge b){
return a.weight < b.weight;
}
int main(){
int n;
while(scanf("%d",&n)!=EOF,n){
int i;
int cnt = 1;
for(i = 1 ; i < n ; ++i){
char start;
int roads;
cin >> start >> roads;
int j;
for(j = 1 ; j <= roads ; ++j){
char end;
int weight;
cin >> end >> weight;
//将字符索引转化成数字索引..
edges[cnt].begin = start - ‘A‘ + 1;
edges[cnt].end = end - ‘A‘ + 1;
edges[cnt++].weight = weight;
}
}
cnt -= 1;
sort(edges+1,edges+1+cnt,cmp);
printf("%d\n",kruscal(cnt));
}
return 0;
}