HDU 1301(MST)

简介: Jungle Roads Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2932 Accepted Submission(s): 2074 ...

Jungle Roads

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2932 Accepted Submission(s): 2074


Problem 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.

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.
 

 

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
 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <iostream>
 4 
 5 using namespace std;
 6 
 7 int u[100], v[100], w[100], p[100], r[100];
 8 int cnt;
 9 
10 int find(int x)
11 {
12     if (x != p[x])
13         p[x] = find(p[x]);
14     return p[x];
15 }
16 
17 int cmp(const int i, const int j)
18 {
19     return w[i] < w[j];
20 }
21 
22 void solve(void)
23 {
24     int ans = 0;
25 
26     for (int i = 0; i <= 30; ++i) p[i] = i;
27     for (int i = 0; i <= cnt; ++i) r[i] = i;
28     sort(r, r + cnt, cmp);//不可去掉,因为是按w[i]排序,不是r[i] 
29     for (int i = 0; i < cnt; ++i) 
30     {
31         int a = find(u[r[i]]);
32         int b = find(v[r[i]]);
33         if (a != b) 
34         {
35             p[a] = b;
36             ans += w[r[i]];
37         }
38     }
39     printf("%d\n", ans);
40 }
41 
42 int main()
43 {
44     int N, num1, num2;
45     char s1[5], s2[5];
46     
47     while (scanf("%d", &N) && N ) 
48     {
49         cnt = 0;
50         for (int i = 1; i <= N - 1; ++i) 
51         {
52             scanf("%s%d", &s1, &num1);
53             if (num1 == 0) continue;
54             for (int j = 1; j <= num1; ++j) 
55             {
56                 scanf("%s%d", s2, &num2);
57                 u[cnt] = s1[0] - 'A';
58                 v[cnt] = s2[0] - 'A';
59                 w[cnt] = num2;
60                 cnt++;
61            }
62         }
63         solve();
64     }
65     
66     return 0;
67 }

 

目录
相关文章
|
7月前
|
机器学习/深度学习 安全 Java
hdu-1596-find the safest road(dijkstra)
hdu-1596-find the safest road(dijkstra)
46 0
hdu 1052 Tian Ji -- The Horse Racing【田忌赛马】(贪心)
hdu 1052 Tian Ji -- The Horse Racing【田忌赛马】(贪心)
58 0
|
算法
POJ3061 Subsequence
POJ3061 Subsequence
|
机器学习/深度学习 人工智能 BI
codeforces-1242-B 0-1 MST
B. 0-1 MST time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Ujan has a lot of useless stuff in his drawers, a considerable part of which are his math notebooks: it is time to sort them out.
168 0
codeforces-1242-B 0-1 MST
[HDU 4738] Caocao‘s Bridges | Tarjan 求割边
Problem Description Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But he wouldn’t give up. Caocao’s army still was not good at water battles, so he came up with another idea. He built many islands in the Changjiang river,
144 0
|
Java 测试技术 C++
HDU 3783 ZOJ
ZOJ Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2779    Accepted Submission(s): 1840 Problem Description 读入一个字符串,字符串中包含ZOJ三个字符,个数不一定相等,按ZOJ的顺序输出,当某个字符用完时,剩下的仍然按照ZOJ的顺序输出。
1110 0