注意：
注意数组越界问题（提交出现runtimeError代表数组越界）
刚开始提交的时候，边集中边的数目和点集中点的数目用的同一个宏定义，但是宏定义是按照点的最大数定义的，所以提交的时候出现了数组越界问题，以后需要注意啦。
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
//数组越界会出现runtimeerror的错误，注意边的数目和点的数目不一样，用一个宏定义的时候注意是否会出现数组越界问题 /* 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：Kruskal算法 */ #include <stdio.h> #include <algorithm> #include <stdlib.h> #include <iostream> using namespace std; const int MAXN=30; /*边结构*/ typedef struct{ int start;//道路起点 int end;//道路终点 double value;//道路权值 }Edge; Edge road[80]; /*节点集合*/ int node[MAXN]; /*寻根函数*/ int Find_set(int n){ if(node[n]==-1) return n; return node[n] == n ? node[n] : Find_set(node[n]); //return node[n]=Find_set(node[n]); } /*排序中的比较函数*/ bool cmp(Edge a,Edge b){ if(a.value<b.value) return true; return false; } /*合并：将棵树合并成一棵树*/ bool Merge(int a,int b){ int r1=Find_set(a); int r2=Find_set(b); if(r1==r2) return false; if(r1<r2) node[r2]=r1; if(r2<r1) node[r1]=r2; return true; } /*克鲁斯卡尔算法*/ int Kruskal(int N,int M){ //N 顶点数 M 边数 int num=0; int cost=0; sort(road,road+M,cmp); for(int i=0;i<M;i++){ if(Merge(road[i].start,road[i].end)){ num++; cost+=road[i].value; } if(num==N-1) break; } if(num!=N-1) return -1; //不能产生最小生成树 else return cost; } int main() { int n; //freopen("input.txt", "r", stdin); while(scanf("%d", &n) != EOF) { if(n == 0) break; for(int i = 0; i < n; i++) node[i] = i; char s, e; int num, cost, k = 0; for(int i = 0; i < n-1; i++) { cin >> s >> num; for(int j = 0; j < num; j++, k++) { cin >> e >> cost; road[k].start = s - 'A'; road[k].end = e - 'A'; road[k].value = cost; } } sort(road, road+k, cmp); int ret = Kruskal(n, k); printf("%d ", ret); } return 0; }