2013
12-04

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.

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

216
30

#include <iostream>
using namespace std;
#include <stdlib.h>
#include <string.h>
typedef struct{
int a,b;
int len;
}Edge;

Edge input[120];

/*********************************************************/
/*并查集                                                 */
/*********************************************************/
int parent[30]; //节点个数
int root(int n){
if(parent[n] == 0) return n;
else return parent[n] = root(parent[n]);
}

/*********************************************************/
/*kruskal                                                */
/*********************************************************/
int cmp(const void  *a,const void  *b){
return (((Edge *)a)->len - ((Edge *)b)->len);
}
int kuruskal(int edge_top){
int a_root,b_root;
int sum = 0;
for(int i = 0; i < edge_top;++i){
a_root = root(input[i].a);
b_root = root(input[i].b);
if(a_root != b_root){
a_root < b_root?parent[b_root] = a_root:parent[a_root] = b_root;
sum += input[i].len;
}
}
return sum;
}
/*********************************************************/
int main(int argc, char *argv[])
{
//FILE *fp;
//fp = freopen("in3.txt","r",stdin);
int n,a_v_Edge_num,edge_top;
char tem,tem_an;

while(scanf("%d%*c",&n),n){
edge_top = 0;

for(int i = 0; i < n - 1; ++i){
scanf("%c %d%*c",&tem,&a_v_Edge_num);
//getchar();
//cin>>tem>>a_v_Edge_num;
while(a_v_Edge_num --){
scanf("%c %d%*c",&tem_an,&input[edge_top].len);
//getchar();
//cin>>tem_an>>input[edge_top].len;

input[edge_top].a = tem - 'A' + 1;
input[edge_top].b = tem_an - 'A' + 1;
++edge_top;
}
}
qsort(input,edge_top,sizeof(Edge),cmp);
memset(parent,0,sizeof(parent));
printf("%d\n",kuruskal(edge_top));
}
return 0;
}

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2. 我没看懂题目
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
我觉得第一个应该是5 6 -1 5 4 输出是19 5 4
第二个是7 0 6 -1 1 -6 7输出是14 7 7
不知道题目例子是怎么得出来的

3. 第一句可以忽略不计了吧。从第二句开始分析，说明这个花色下的所有牌都会在其它里面出现，那么还剩下♠️和♦️。第三句，可以排除2和7，因为在两种花色里有。现在是第四句，因为♠️还剩下多个，只有是♦️B才能知道答案。