UVA 11729 - Commando War

本文介绍了一个基于任务执行时间和简报时间的士兵任务调度问题。通过分析每个士兵的任务完成时间及简报所需时间,采用特定算法确保所有任务以最短总时间完成。该问题通过排序和迭代计算每一步的最佳顺序来解决。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >



G

Commando War

Input: Standard Input

Output: Standard Output

 

 

“Waiting for orders we held in the wood, word from the front never came

By evening the sound of the gunfire was miles away

Ah softly we moved through the shadows, slip away through the trees

Crossing their lines in the mists in the fields on our hands and our knees

And all that I ever, was able to see

The fire in the air, glowing red, silhouetting the smoke on the breeze”

 

There is a war and it doesn't look very promising for your country. Now it's time to act. You have a commando squad at your disposal and planning an ambush on an important enemy camp located nearby. You have soldiers in your squad. In your master-plan, every single soldier has a unique responsibility and you don't want any of your soldier to know the plan for other soldiers so that everyone can focus on his task only. In order to enforce this, you brief every individual soldier about his tasks separately and just before sending him to the battlefield. You know that every single soldier needs a certain amount of time to execute his job. You also know very clearly how much time you need to brief every single soldier. Being anxious to finish the total operation as soon as possible, you need to find an order of briefing your soldiers that will minimize the time necessary for all the soldiers to complete their tasks. You may assume that, no soldier has a plan that depends on the tasks of his fellows. In other words, once a soldier  begins a task, he can finish it without the necessity of pausing in between.

 

Input

 

There will be multiple test cases in the input file. Every test case starts with an integer N (1<=N<=1000), denoting the number of soldiers. Each of the following N lines describe a soldier with two integers B (1<=B<=10000) J (1<=J<=10000)seconds are needed to brief the soldier while completing his job needs seconds. The end of input will be denoted by a case with N =0 . This case should not be processed.

 

Output

 

For each test case, print a line in the format, “Case X: Y”, where X is the case number & Y is the total number of seconds counted from the start of your first briefing till the completion of all jobs.

 


#include <stdio.h>

#include <math.h>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
struct soldiers
{
    int b;
    int job;
}s[1010];
bool cmp(soldiers s1,soldiers s2)
{
    return (s1.job)>(s2.job);
}
int main()
{
    int n;
    int kcase=0;
    while(scanf("%d",&n)==1)
    {
        kcase++;
        if(n==0)
        break;
        int i;
        for(i = 0;i<n;++i)
        scanf("%d%d",&s[i].b,&s[i].job);
        sort(s,s+n,cmp);
        int sum=0;
        int maxn=s[0].b+s[0].job;
        for(int j = 1;j<n;++j)
        {
         for(i = 0;i<=j;++i)
         {
         sum+=s[i].b;
         }
         sum+=s[j].job;
         maxn=max(maxn,sum);
         sum=0;
        }
        printf("Case %d: %d\n",kcase,maxn);
    }
    return 0;
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值