Jungle Roads
2016-07-03 12:17
375 查看
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
Sample Output
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int inf = 0x3fffffff;
int p[27];
struct prog {
int u;
int v;
int w;
}mymap[80];
bool cmp ( prog a , prog b)
{
return a.w<b.w;
}
int myfind(int x)
{
return x==p[x]?x:p[x]=myfind(p[x]);
}
int main()
{
int n;
while ( cin >> n , n )
{
int i , j ;
for ( i = 0 ; i < 27 ; i ++ )
p[i] = i ;
int k = 0 ;
for ( i = 0 ; i < n - 1 ; i ++ )
{
char str[3];
int m;
cin >> str >> m ;
for ( j = 0 ; j < m ; j ++ ,k ++ )
{
char str2[3];
int t;
cin >> str2 >> t ;
mymap[k].u=(str[0]-'A');
mymap[k].v=(str2[0]-'A');
mymap[k].w=t;
}
}
sort ( mymap , mymap + k , cmp );
int ans=0;
for ( i = 0 ; i < k ; i ++ )
{
int x = myfind(mymap[i].u);
int y = myfind(mymap[i].v);
if( x!=y)
{
ans+=mymap[i].w;
p[x]=y;
}
}
cout<<ans<<endl;
}
return 0;
}
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 题意:所有的桥都坏了,需要重修,每个桥修的时间不一样,求所有的桥都链接上的最短时间。 思路:求的是最小生成树,生成树有两种算法,这边使用Kruskal,其算法思想
假设WN=(V,{E})是一个含有n个顶点的连通网,则按照克鲁斯卡尔算法构造最小生成树的过程为: 先构造一个只含n个顶点,而边集为空的子图, 若将该子图中各个顶点看成是各棵树上的根结点,则它是一个含有n棵树的一个森林。 之后,从网的边集E中选取一条权值最小的边,若该条边的两个顶点分属不同的树,则将其加入子图, 也就是说,将这两个顶点分别所在的两棵树合成一棵树; 反之,若该条边的两个顶点已落在同一棵树上,则不可取,而应该取下一条权值最小的边再试之。 依次类推,直至森林中只有一棵树,也即子图中含有n-1条边为止。 代码:
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int inf = 0x3fffffff;
int p[27];
struct prog {
int u;
int v;
int w;
}mymap[80];
bool cmp ( prog a , prog b)
{
return a.w<b.w;
}
int myfind(int x)
{
return x==p[x]?x:p[x]=myfind(p[x]);
}
int main()
{
int n;
while ( cin >> n , n )
{
int i , j ;
for ( i = 0 ; i < 27 ; i ++ )
p[i] = i ;
int k = 0 ;
for ( i = 0 ; i < n - 1 ; i ++ )
{
char str[3];
int m;
cin >> str >> m ;
for ( j = 0 ; j < m ; j ++ ,k ++ )
{
char str2[3];
int t;
cin >> str2 >> t ;
mymap[k].u=(str[0]-'A');
mymap[k].v=(str2[0]-'A');
mymap[k].w=t;
}
}
sort ( mymap , mymap + k , cmp );
int ans=0;
for ( i = 0 ; i < k ; i ++ )
{
int x = myfind(mymap[i].u);
int y = myfind(mymap[i].v);
if( x!=y)
{
ans+=mymap[i].w;
p[x]=y;
}
}
cout<<ans<<endl;
}
return 0;
}
相关文章推荐
- Vue.js——60分钟组件快速入门(下篇)
- mybatis参数映射
- 编译安装 Centos 7 x64 + tengine.2.0.3 (实测+笔记)
- NPM 使用介绍
- 使用Intent传递对象的两种方法(Serializable,Parcelable)
- java web
- Yii2 使用 Joins 查询
- opencv学习(3)——addWeighted函数将两幅图像叠加
- UDP编程中的connect
- VM 下安装ghost版系统
- Behavior Designer中Wait节点的坑
- [置顶] 课程设计 --- 黑白棋中的 AI
- java web
- C++单链表基本操作练习
- 记一次阿里云服务器发布
- c标准基本库函数:libc、glibc和glib的关系
- Hive是什么?
- PS利用自由变换制作飞舞的蝴蝶
- Fedora 22/23升级到 Fedora 24
- spring aop之对象内部方法间的嵌套失效