POJ 1251 Jungle Roads(Prim算法)
来源:互联网 发布:风油精滴到私处知乎 编辑:程序博客网 时间:2024/05/21 17:33
Jungle Roads
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 18 Accepted Submission(s) : 14
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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
9A 2 B 12 I 25B 3 C 10 H 40 I 8C 2 D 18 G 55D 1 E 44E 2 F 60 G 38F 0G 1 H 35H 1 I 353A 2 B 10 C 40B 1 C 200
Sample Output
21630
题目读懂以后这道题目就是最直接的求最小生成树
除了建图比模板图复杂一点别的都一样
AC:
#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std;
int map[30][30];
int dist[30];
int visited[30];
int n,t;
const int inf=99999;
void prim()
{
int min,pos,sum=0;
memset(visited,0,sizeof(visited));
for(int i=1;i<=t;i++)
dist[i]=map[1][i];
visited[1]=1;
for(int i=1;i<t;i++)
{
min=inf;
for(int j=1;j<=t;j++)
{
if(!visited[j]&&dist[j]<min)
{
min=dist[j];
pos=j;
}
}
if(min==inf) break;
visited[pos]=1;
sum+=min;
for(int j=1;j<=t;j++)
{
if(!visited[j]&&map[pos][j]<dist[j])
dist[j]=map[pos][j];
}
}
cout<<sum<<endl;
}
int main()
{
int k,g;
char str[3];
char str2[3];
while(~scanf("%d",&t),t)
{
for(int i=1;i<=t;i++)
for(int j=1;j<=t;j++)
{
map[i][j]=inf;
if(i==j) map[i][j]=0;
}
for(int i=1;i<t;i++)
{
scanf("%s %d",str,&k);
while(k--)
{
scanf("%s %d",str2,&g);
map[str[0]-'A'+1][str2[0]-'A'+1]=g;
map[str2[0]-'A'+1][str[0]-'A'+1]=g;
}
}
prim();
}
return 0;
}
#include<cstdio>
#include<string.h>
using namespace std;
int map[30][30];
int dist[30];
int visited[30];
int n,t;
const int inf=99999;
void prim()
{
int min,pos,sum=0;
memset(visited,0,sizeof(visited));
for(int i=1;i<=t;i++)
dist[i]=map[1][i];
visited[1]=1;
for(int i=1;i<t;i++)
{
min=inf;
for(int j=1;j<=t;j++)
{
if(!visited[j]&&dist[j]<min)
{
min=dist[j];
pos=j;
}
}
if(min==inf) break;
visited[pos]=1;
sum+=min;
for(int j=1;j<=t;j++)
{
if(!visited[j]&&map[pos][j]<dist[j])
dist[j]=map[pos][j];
}
}
cout<<sum<<endl;
}
int main()
{
int k,g;
char str[3];
char str2[3];
while(~scanf("%d",&t),t)
{
for(int i=1;i<=t;i++)
for(int j=1;j<=t;j++)
{
map[i][j]=inf;
if(i==j) map[i][j]=0;
}
for(int i=1;i<t;i++)
{
scanf("%s %d",str,&k);
while(k--)
{
scanf("%s %d",str2,&g);
map[str[0]-'A'+1][str2[0]-'A'+1]=g;
map[str2[0]-'A'+1][str[0]-'A'+1]=g;
}
}
prim();
}
return 0;
}
0 0
- POJ 1251 Jungle Roads(Prim算法)
- poj 1251 Jungle Roads(Prim)
- poj 1251 Jungle Roads(Prim)
- POJ - 1251----Jungle Roads(prim)
- poj 1251 Jungle Roads 【prim】
- POJ 1251 Jungle Roads (prim)
- POJ 1251 Jungle Roads (Prim算法)&& (Kruskal算法)
- POj 1251 Jungle Roads (Kruskal算法 + Prim算法)
- POJ 1251 Jungle Roads (最小生成树 Prim普里姆算法)
- POJ-1251 Jungle Roads 最小生成树(prim算法)
- POJ 1251 Jungle Roads ( Prim , Kruscal )
- poj 1251 Jungle Roads【prim & kruskal】
- POJ1251 Jungle Roads Prim算法
- poj 1251 Jungle Roads prim最小生成树基础
- POJ 1251 Jungle Roads(最小生成树裸prim)
- hdu 1301 Jungle Roads &&poj 1251 (prim 或者kruskal)
- POJ 1251 Jungle Roads(Prim or kruskal)
- poj 1251 Jungle Roads 解题报告(kruskal+prim)
- 面向对象特征
- [又值奥运季] 世界各国国旗与地图对照
- GC的前世与今生
- Jstorm2.1.1集群安装
- STL vector deque list set map的erase()
- POJ 1251 Jungle Roads(Prim算法)
- BZOJ3680 吊打XXX
- opencv3.1.0 特征点检测与图像匹配(features2d、xfeatures2d)
- 使用pbc
- 洛谷 P2330 05四川 繁忙的都市
- C语言 字符数组转换整数 stringToInt
- 《TCP/IP详解 卷1:协议》 读书笔记 第3章 IP:网际协议
- 平面方程推导
- 华为上机题汇总(一)