博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
hdu 1301&&poj 1251 最小生成树prim实现
阅读量:4967 次
发布时间:2019-06-12

本文共 3944 字,大约阅读时间需要 13 分钟。

http://acm.hdu.edu.cn/showproblem.php?pid=1301

Jungle Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4145    Accepted Submission(s): 3020

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

 

 

 

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

经典的prim算法,但是要注意存入图的方式,并且要注意是无向图,赋值要对称赋值

#include 
#include
#include
#include
#define Maxint 0x7fffffffint map[105][105],visit[105],low[105];int n;int prim(){ int pos=1,i,j,minn,sum=0; memset(visit,0,sizeof(visit)); visit[1]=1; for(i=1;i<=n;i++) { if(i!=pos) { low[i]=map[pos][i]; } } for(i=1;i
low[j]) { minn=low[j]; pos=j; } } sum+=minn; visit[pos]=1; for(j=1;j<=n;j++) { if(visit[j]==0 && low[j]>map[pos][j]) { low[j]=map[pos][j]; } } } return sum;}int main(){ int i,j; while(scanf("%d",&n)!=EOF&&n!=0) { char a,b,c; int m,k; for(i=1;i<=n;i++) for(j=1;j<=n;j++) map[i][j]=Maxint; getchar(); for(i=1;i

  

 //poj1251 注意空格,应改用cin输入,以防止多个空格造成PE

转载于:https://www.cnblogs.com/ccccnzb/p/3832301.html

你可能感兴趣的文章
表单校验之datatype
查看>>
python第六篇文件处理类型
查看>>
ubuntu16系统磁盘空间/dev/vda1占用满的问题
查看>>
grid网格布局
查看>>
JSP常用标签
查看>>
dashucoding记录2019.6.7
查看>>
九涯的第一次
查看>>
处理器管理与进程调度
查看>>
页面懒加载
查看>>
向量非零元素个数_向量范数详解+代码实现
查看>>
java if 用法详解_Java编程中的条件判断之if语句的用法详解
查看>>
java -f_java学习笔记(一)
查看>>
java 什么题目好做_用java做这些题目
查看>>
java中的合同打印_比较方法违反了Java 7中的一般合同
查看>>
php 位运算与权限,怎么在PHP中使用位运算对网站的权限进行管理
查看>>
matlab sin函数 fft,matlab的fft函数的使用教程
查看>>
wcdma下行如何解扩解扰 matlab,WCDMA技术基础.ppt
查看>>
mysql adddate()函数
查看>>
mysql 根据日期时间查询数据
查看>>
mysql sin() 函数
查看>>