最小生成树Jungle Roads

这道题一定要注意录入方式，我用的解法是prime算法

因为单个字符的录入会涉及到缓冲区遗留的空格问题，我原本是采用c语言的输入方法录入数据的，结果对了，但是提交却一直wrong，后来改成了c++的cin方法来录入，就对了

先把提交对的代码和提交错的代码都放上来，欢迎提出建议

prime算法：

　　1：清空生成树，任取一个顶点加入生成树

　　2：在那些一个顶点在生成树中，一个顶点不在生成树的边中，选取一条权最小的边，将他和另一个端点加入生成树

　　3：重复步骤2，直到所有的点都进入了生成树为止，此时的生成树就是最小生成树。

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

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAXN 30
#define inf 1000000000
typedef int elem_t;
/*
    题意：
        多组案例
        每组案例第一行输入一个数字n
        下面n-1行
        每行的第一个数据都是一个字符start，字符从A往后依次排列
        每行的第二个数据是一个数字num，表示有num个节点与该行第一个字符表示的节点相连
        每行接下来的数据是num组end,cost，表示start到end的花费为cost
        具体输入输出看案例就会懂
    解法：prime算法
*/
elem_t prim(int n, elem_t mat[][MAXN], int* pre)
{
    elem_t min[MAXN], ret = ;
    int v[MAXN], i, j, k;
    for(i = ; i < n; i++)
        min[i] = inf, v[i] = , pre[i] = -;
    for(min[j = ] = ; j < n; j++)
    {
        for(k = -, i = ; i < n; i++)
            if(!v[i] && (k == - || min[i] < min[k]))
                k = i;
        for(v[k] = , ret += min[k], i = ; i < n; i++)
            if(!v[i] && mat[k][i] < min[i])
                min[i] = mat[pre[i] = k][i];
    }
    return ret;
}
int main()
{
    int n;
    elem_t mat[MAXN][MAXN];
    int pre[MAXN];
    //freopen("input.txt", "r", stdin);
    while(scanf("%d", &n) != EOF)
    {
        if(n == )
            break;
        for(int i = ; i < n; i++)
            for(int j = ; j <= i; j++)
                mat[i][j] = mat[j][i] = inf;
        char s, e;
        int num, cost;
        for(int i = ; i < n-; i++)
        {
            cin >> s >> num;
            for(int j = ; j < num; j++)
            {
                cin >> e >> cost;
                mat[s-'A'][e-'A'] = mat[e-'A'][s-'A'] = cost;
            }
        }
        elem_t mi = prim(n, mat, pre);
        printf("%d\n", mi);
    }
    return ;
}

下面是没过的代码

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define MAXN 30
#define inf 1000000000
typedef int elem_t;
elem_t prim(int n, elem_t mat[][MAXN], int* pre)
{
    elem_t min[MAXN], ret = ;
    int v[MAXN], i, j, k;
    for(i = ; i < n; i++)
        min[i] = inf, v[i] = , pre[i] = -;
    for(min[j = ] = ; j < n; j++)
    {
        for(k = -, i = ; i < n; i++)
            if(!v[i] && (k == - || min[i] < min[k]))
                k = i;
        for(v[k] = , ret += min[k], i = ; i < n; i++)
            if(!v[i] && mat[k][i] < min[i])
                min[i] = mat[pre[i] = k][i];
    }
    return ret;
}
int main()
{
    int n;
    elem_t mat[MAXN][MAXN];
    int pre[MAXN];
    freopen("input.txt", "r", stdin);
    while(scanf("%d", &n) != EOF)
    {
        if(n == )
            break;
        for(int i = ; i < n; i++)
        {
            for(int j = ; j < n; j++)
                mat[i][j] = inf;
        }
        for(int i = ; i < n-; i++)
        {
            char ch1, ch2;
            getchar();
            scanf("%c", &ch1);
            int num;
            elem_t cost;
            scanf("%d", &num);
            for(int j = ; j < num; j++)
            {
                scanf(" %c%d", &ch2, &cost);
                if(cost < mat[ch1-'A'][ch2-'A'])
                    mat[ch1-'A'][ch2-'A'] = mat[ch2-'A'][ch1-'A'] = cost;
            }
        }
        elem_t mi = prim(n, mat, pre);
        printf("%d\n", mi);
    }
    return ;
}