POJ-1251（最小生成树）

Jungle Roads

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22634		Accepted: 10545

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. 

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 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

题意：有n个城市，有好多条路，现在那个国家穷了，需要去掉一些路来减少维修费用。输入，先输入n，代表n个城市。一下n-1行分别先输入一个字符c1，代表城市的标号。后边紧接着的一个数字代表该城市的边。后边每一对字符c2和数字w代表c1到c2有路，维修费是w。
思路：直接克鲁斯卡尔。

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;

#define maxm 1000
int n, k;
//u，v，w的数组小标表示边的序号。
//u[i]表示第i条边的u点，v[i]表示第i条边的v点，w[i]表示第i条边的w点
int u[maxm], v[maxm], w[maxm];
int p[maxm];//并查集的数组
int r[maxm];//按照边权排序后的边号

//比较函数，按边权从小到大排序。
int cmp(const int i, const int j){
    return w[i] < w[j];
}

//并查集find函数
int find(int x){
    return p[x] == x ? x : p[x] = find(p[x]);//带压缩路径的
}

int kru(){
    //初始化
    int ans = 0;
    for (int i = 0; i < n; i++) p[i] = i;
    for (int i = 0; i < k; i++) r[i] = i;
    sort(r, r + k, cmp);

    for (int i = 0; i < k; i++){//按照权值从小到大遍历所有边
        int e = r[i];
        int x = find(u[e]);//检查边的u点v点是否处在同一连通分量
        int y = find(v[e]);
        if(x != y){//如果不在同一连通分量，则合并
            ans += w[e];//计入权值和
            p[x] = y;
        }
    }
    return ans;
}

int main()
{
    while(cin >> n){
        if(n == 0) break;
        k = 0;//记录边的总数
        for(int i = 0; i < n-1; i++){
            char c1;
            int num1;
            cin >> c1 >> num1;
            for (int j = 0; j < num1; j++){
                char c2;
                int num2;
                cin >> c2 >> num2;
                u[k] = c1-'A';//将对应的城市转换成点
                v[k] = c2-'A';
                w[k++] = num2;
            }
        }
        printf("%d\n", kru());
    }
    return 0;
}