hdu 1301&& ny 434 &&poj 1251 Jungle Roads【最小生成树】
2015-08-11 20:55
711 查看
Jungle Roads
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5454 Accepted Submission(s): 3935
[align=left]Problem Description[/align]
![](http://acm.hdu.edu.cn/data/images/1301-1.gif)
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.
[align=left]Sample Input[/align]
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
[align=left]Sample Output[/align]
216
30
题意:
给出一些编号为大写字母的城市,A 表示第一个城市,依次往后,最多26个城市,某些城市之间有很多道路相连,现在给出所有的城市,以及他们之间的道路的数量长度,让你从这些道路中找到最短的路程,让所有的城市都连通,输出最短的道路长度。
比较简单的最小生成树模板题目,只需要控制好格式,其他地方问题不大,克鲁斯卡尔算法比较适合这一类的题目,注意条件的控制。
这个题的输入特别坑,不知道为什么。在poj和南阳上,如果用getchar来吸收换行,那么会显示re,没办法了,只能在输入前面加空格来吸收了....
#include<stdio.h> #include<algorithm> #include<string.h> using namespace std; int n,per[30],cnt,kase; struct road { int a,b; int len; }x[105]; void init()//初始化 { for(int i=1;i<=n;++i) { per[i]=i; } } int find(int x)//查找 { int r=x; while(r!=per[r]) { r=per[r]; } int i=x,j; while(i!=r) { j=per[i];per[i]=r;i=j; } return r; } void join(int a,int b)//合并 { int fx=find(a),fy=find(b); if(fx!=fy) { per[fy]=fx; ++cnt;kase=1;//统计边数 } } int cmp(road a,road b) { return a.len<b.len; } int main() { int i,j,m,a,c;char u,v; while(~scanf("%d",&n),n) { init();c=cnt=0; for(i=0;i<n-1;++i) { scanf(" %c %d",&u,&m); for(j=0;j<m;++j) { scanf(" %c %d",&v,&a); x[c].a=u-'A'+1;x[c].b=v-'A'+1;x[c].len=a;//把对应的字母转化为数字存下来 ++c;//累加道路数量 } getchar(); } sort(x,x+c,cmp);//排序 int sum=0; for(i=0;i<c&&cnt<n-1;++i) { kase=0; join(x[i].a,x[i].b); if(kase)//是否加入成功 { sum+=x[i].len; } } printf("%d\n",sum); } return 0; }
相关文章推荐
- Tomcat 设计模式总结(Tomcat源代码阅读系列之八)
- Android自定义属性时TypedArray的使用方法
- 1- Two Sum
- Excel中的数据导入到SqlServer数据库中
- 用链表实现电子词典
- win7上配置基于CUDA计算的Theano深度学习环境
- iOS基础-UIKit框架-高级视图-UIPickerView-实例3:国家选择(图片)
- 简单解决java.lang.IllegalStateException: getOutputStream() has already been called for this
- exception in thread main brut.androlib.err.undefinedresobject resource spec : 0x01010462
- 类的复制构造函数调用
- 前言:关于nagios监控
- php时区 为什么两个语句才能起效 date_default_timezone_set ini_set('date.timezone
- 8.11(web doget方法乱码)
- 前言:关于nagios监控
- object转化为string
- Linux后门入侵检测方法以及工具
- 九度oj 1007
- 教你自己实现一个事件总线EventBus
- 寻找数组中的第二大数
- OC_AddressBook_通讯录写入