POJ1251 Jungle Roads Kruskal+scanf输入小技巧
Jungle Roads
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
Output
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 题意:经典最小生成树。
思路:有向边无向边对生成树来说不重要。顺序读入即可。这里有个小技巧,scanf里的空格可以吃掉连续一串(可以为空)空白符(空格,tab,换行),直到出现第一个非空白符,比getchar()更适用。
#include<stdio.h>
#include<algorithm>
using namespace std; int f[],a[];
struct Edge{
int u,v,w;
}edge[]; bool cmp(Edge a,Edge b)
{
return a.w<b.w;
} int find(int x)
{
return f[x]==x?x:f[x]=find(f[x]);
} int kru(int c,int n)
{
int i;
for(i=;i<=n;i++){
f[i]=i;
}
sort(edge+,edge+c+,cmp);
int cnt=,ans=;
for(i=;i<=c;i++){
int u=edge[i].u;
int v=edge[i].v;
int w=edge[i].w;
int fu=find(u),fv=find(v);
if(fu!=fv){
ans+=w;
f[fv]=fu;
cnt++;
}
if(cnt==n-) break;
}
if(cnt<n-) return -;
else return ans;
}
int main()
{
int n,x,t,c,i,j;
char ch1,ch2;
while(scanf("%d",&n)&&n!=){
c=;
for(i=;i<n;i++){
scanf(" %c %d",&ch1,&t); //
for(j=;j<=t;j++){
scanf(" %c %d",&ch2,&x); //getchar()个别出现RE情况
edge[++c].u=ch1-'A'+;
edge[c].v=ch2-'A'+;
edge[c].w=x;
}
}
printf("%d\n",kru(c,n));
}
return ;
}
POJ1251 Jungle Roads Kruskal+scanf输入小技巧的更多相关文章
- POJ1251 Jungle Roads(Kruskal)(并查集)
Jungle Roads Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 23882 Accepted: 11193 De ...
- poj1251 Jungle Roads Kruskal算法+并查集
时限: 1000MS 内存限制: 10000K 提交总数: 37001 接受: 17398 描述 热带岛屿拉格里山的首长有个问题.几年前,大量的外援花在了村庄之间的额外道路上.但是丛林不断地超 ...
- HDU1301&&POJ1251 Jungle Roads 2017-04-12 23:27 40人阅读 评论(0) 收藏
Jungle Roads Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 25993 Accepted: 12181 De ...
- POJ1251 Jungle Roads 【最小生成树Prim】
Jungle Roads Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 19536 Accepted: 8970 Des ...
- POJ1251 Jungle Roads (最小生成树&Kruskal&Prim)题解
题意: 输入n,然后接下来有n-1行表示边的加边的权值情况.如A 2 B 12 I 25 表示A有两个邻点,B和I,A-B权值是12,A-I权值是25.求连接这棵树的最小权值. 思路: 一开始是在做莫 ...
- poj1251 Jungle Roads(Prime || Kruskal)
题目链接 http://poj.org/problem?id=1251 题意 有n个村庄,村庄之间有道路连接,求一条最短的路径能够连接起所有村庄,输出这条最短路径的长度. 思路 最小生成树问题,使用普 ...
- POJ1251 Jungle Roads【最小生成树】
题意: 首先给你一个图,需要你求出最小生成树,首先输入n个节点,用大写字母表示各节点,接着说有几个点和它相连,然后给出节点与节点之间的权值.拿第二个样例举例:比如有3个节点,然后接下来有3-1行表示了 ...
- HDU1301 Jungle Roads(Kruskal)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission( ...
- POJ1251 Jungle Roads
解题思路:看懂题意是关键,Kruskal算法,最小生成树模板. 上代码: #include<cstdio> #include<cstring> #include<algo ...
随机推荐
- c++中的重载、覆盖和隐藏
1 重载发生在同一个类内部. 同一个类内部,具有相同的函数名,但是参数列表不同,那么就是重载.因为c++编译器编译时,将函数名和函数列表一起对函数进行了重命名. 2 覆盖和隐藏发生在子类和父类之间. ...
- iOS与H5交互及UIWebView缓存
iOS原生App与H5页面交互笔记 最近在做一个项目用到了原生App与H5交互,之前有做过简单的H5页面直接调用原生方法的例子,就是利用UIWebView中的代理方法 //webview每次加载之前都 ...
- MapReduce-PRODUCTION-DEMAND
[粗暴的HIVE-SQL]select xyz from abc where ty='sdk' and ret_code=0 and data_source_type=1 and dt between ...
- js实现随机选取[10,100)中的10个整数,存入一个数组,并排序。 另考虑(10,100]和[10,100]两种情况。
1.js实现随机选取[10,100)中的10个整数,存入一个数组,并排序. <!DOCTYPE html> <html lang="en"> <hea ...
- When Programmers and Testers Collaborate
When Programmers and Testers Collaborate Janet Gregory SOMETHING MAGICAL HAPPENS when testers and pr ...
- install_driver(mysql) failed
安装好了mysql监控神器innotop,正得意,innotoop不可用,其错误提示为install_driver(mysql) failed: Can't load '/usr/lib64/ ...
- mapreduce的一个模版
import java.io.IOException; import java.text.DateFormat; import java.text.SimpleDateFormat; import j ...
- 百度小程序转换微信小程序
Python脚本,一键转换Github地址:https://github.com/DWmelon/py-transfer-BdToWx 运行条件 具备Python环境,可在命令行中使用Python命令 ...
- CSS相对|绝对(relative/absolute)定位
1.破坏性和包裹性 position:absolute与float:left,两者有两大共性:包裹性,破坏性. 包裹性 包裹性换种说法就是让元素inline-block化,例如一个div标签默认宽度是 ...
- Java基础知识整理之static修饰属性
static 关键字,我们在开发用的还是比较多的.在<Java编程思想>有下面一段话 static 方法就是没有 this 的方法.在 static 方法内部不能调用非静态方法,反过来是可 ...