poj-1700 crossing river(贪心题)
qq群名繁体字,魔力无双,田欣仪
题目描述:
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
题目大意:过河,速度取慢的
题解:有点像汉诺塔的感觉,贪心算法,直接看代码吧,大水题
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstdlib>
#include<queue>
using namespace std;
#define pi 3.14159265358979323846264338327950
int t,n,a[1003],t;
int main()
{
scanf("%d",&t);
while(t--)
{
t=0;
scanf("%d",&n);
memset(a,0,sizeof(a));
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
if(n==1)
t=a[0];
else if(n==2)
t=a[1];
else if(n==3)
t=a[0]+a[1]+a[2];
else
{
for(int i=n-1;i>0;i=i-2)
{
if(i>=3)
{
if(a[0]+a[i-1]>=2*a[1])
t=t+a[0]+2*a[1]+a[i];
else
t=t+2*a[0]+a[i]+a[i-1];
}
if(i==2)
t=t+a[0]+a[1]+a[2];
if(i==1)
t=t+a[1];
}
}
printf("%d\n",t);
}
}
如对本文有疑问,请在下面进行留言讨论,广大热心网友会与你互动!!
点击进行留言回复
相关文章:
-
-
C++ 作用域
作用域:名称在翻译单元(包括文件)的可见范围 局部: 只在定义它的代码块中可用,如自动变量 全局(文件作用域): 从定义位置到文件结尾都可用 注意: 静...
[阅读全文]
-
-
-
-
聚合类型与POD类型
Lippman在《深度探索C++对象模型》的前言中写道: I have heard a number of people over the years ...
[阅读全文]
-
-
-
-
-
网友评论