POJ 1700 F(Contest #3)
2015-07-19 21:14
169 查看
Description
A group of N people wishes to go across a river with only one boat, which can at most carry two persons. Therefore some sort of shuttle arrangement must be arranged in order to row the boat back and forth so that all people may cross. Each person has a different rowing speed; the speed of a couple is determined by the speed of the slower one. Your job is to determine a strategy that minimizes the time for these people to get across.
Input
The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. The first line of each case contains N, and the second line contains N integers giving the time for each people to cross the river. Each case is preceded by a blank line. There won't be more than 1000 people and nobody takes more than 100 seconds to cross.
Output
For each test case, print a line containing the total number of seconds required for all the N people to cross the river.
Sample Input
Sample Output
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
freopen("in.txt","r",stdin);
int speed[1000];
int t,n,time;
cin>>t;
int i,j;
for(i=0;i<t;i++)
{
cin>>n;
memset(speed,0,sizeof(speed));
for(j=0;j<n;j++)cin>>speed[j];
sort(speed,speed+n); //从小到大排序
time=0;
while(n>3)
{
if(speed[1]*2 > speed[0]+speed[n-2])
time=time+speed[0]+speed[0]+speed[n-1]+speed[n-2];
else time=time+speed[1]+speed[0]+speed[n-1]+speed[1];
n=n-2;
}
if(n==3)
time=time+speed[0]+speed[1]+speed[2];
else if(n==2)
time=time+speed[1];
else if(n==1)
time=time+speed[0];
cout<<time<<endl;
}
return 0;
}
A group of N people wishes to go across a river with only one boat, which can at most carry two persons. Therefore some sort of shuttle arrangement must be arranged in order to row the boat back and forth so that all people may cross. Each person has a different rowing speed; the speed of a couple is determined by the speed of the slower one. Your job is to determine a strategy that minimizes the time for these people to get across.
Input
The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. The first line of each case contains N, and the second line contains N integers giving the time for each people to cross the river. Each case is preceded by a blank line. There won't be more than 1000 people and nobody takes more than 100 seconds to cross.
Output
For each test case, print a line containing the total number of seconds required for all the N people to cross the river.
Sample Input
1 4 1 2 5 10
Sample Output
17
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
freopen("in.txt","r",stdin);
int speed[1000];
int t,n,time;
cin>>t;
int i,j;
for(i=0;i<t;i++)
{
cin>>n;
memset(speed,0,sizeof(speed));
for(j=0;j<n;j++)cin>>speed[j];
sort(speed,speed+n); //从小到大排序
time=0;
while(n>3)
{
if(speed[1]*2 > speed[0]+speed[n-2])
time=time+speed[0]+speed[0]+speed[n-1]+speed[n-2];
else time=time+speed[1]+speed[0]+speed[n-1]+speed[1];
n=n-2;
}
if(n==3)
time=time+speed[0]+speed[1]+speed[2];
else if(n==2)
time=time+speed[1];
else if(n==1)
time=time+speed[0];
cout<<time<<endl;
}
return 0;
}
相关文章推荐
- 使用IntelliJ IDEA编写Scala在Spark中运行
- HDU 3998 Sequence
- WIN7 64位系统Delphi6安装EhLib【向程序发送命令时出现问题】
- 【LeetCode】House Robber I & II 解题报告
- 如何监控第三方应用程序(SOAP or RESTful client)访问HTTPS当数据站点?
- hdu 2519 求组合数
- Redis 源码分析系列1-main函数相关调用分析
- 利用多核多线程进行程序优化
- 欢迎使用CSDN-markdown编辑器
- 《慕客网:IOS基础入门之Foundation框架初体验》学习笔记 <一>
- Keil中添加自己的头文件
- android网络通信------UDP
- Hadoop的第一个程序 wordcount
- Thinphp模板替换
- JavaScript
- HDU 4627 E(Contest #3)
- LeeCode-Same Tree
- 欢迎使用CSDN-markdown编辑器
- 计蒜之道 初赛 第一场 题解 dp 高效 网络流 最小割 最大流 ISAP 模板
- WCF 设计和实现服务协定(01)