zoukankan      html  css  js  c++  java
  • POJ-1700 Crossing River

    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

    1
    4
    1 2 5 10
    

    Sample Output

    17
    对于n<4的情况我们可以直接计算出来。
    对于大于4我们有每次都有两种策略。
    ①最快和次快过去,最快回;最慢和次慢过去,次快回,t=s[2]+s[1]+s[n]+s[2]。
    ②最快和最慢过去,最快回;次慢和最快过去,最快回,t=s[n]+s[0]+s[n-1]+s[0]。选择两者中用时较少的一个策略执行。
    每次决策后我们都发现过去了两个人。
    于是n-=2;
    再次循环。直到n<4;

    代码:
     1 #include<iostream>
     2 #include<stdio.h>
     3 #include<string.h>
     4 #include<algorithm>
     5 using namespace std;
     6 int a[1200];
     7 int main(){
     8     int T;
     9     cin>>T;
    10     while(T--){
    11         int n;
    12         cin>>n;
    13         for(int i=1;i<=n;i++){
    14             scanf("%d",a+i);
    15         }
    16         sort(a+1,a+1+n);
    17         int sum=0;
    18         while(n){
    19             if(n==1){
    20                 sum+=a[1];
    21                 break;
    22             }
    23             else if(n==2){
    24                 sum+=a[2];
    25                 break;
    26             }
    27             else if(n==3){
    28                 sum+=a[1]+a[2]+a[3];
    29                 break;
    30             }
    31             else{
    32                 sum+=min(a[2]+a[1]+a[n]+a[2],a[n]+a[1]+a[n-1]+a[1]);
    33                 n-=2;
    34             }
    35         }
    36         cout<<sum<<endl;
    37     }
    38     return 0;
    39 }
  • 相关阅读:
    Error: listen EADDRINUSE :::3000
    getElementsByTagName
    AWTK是如何保证代码质量的
    $("div:has(p)") 选取含有元素的元素
    $("div:contains('我')选取含有文本"我"的元素
    不怕从头再来,他通过种植黄连发家致富
    生活所迫下的创业,靠空气清新机的一笔订单他就赚了100万元
    她在创业过程中,将员工转变为合伙人
    数据开源工具:Hadoop为企业带来什么?
    团队中的 Node.js 具体实践
  • 原文地址:https://www.cnblogs.com/ISGuXing/p/7306077.html
Copyright © 2011-2022 走看看