UVA - 11729 Commando War

Commando War

Time Limit: 1000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

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 N 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). B seconds are needed to brief the soldier while completing his job needs J 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.

 

Sample Input����������������������������������������������Output for Sample Input

3 2 5 3 2 2 1 3 3 3 4 4 5 5 0

Case 1: 8 Case 2: 15

---

Problem Setter: Mohammad Mahmudur Rahman, Special Thanks: Manzurur Rahman Khan

Source

Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Problem Solving Paradigms :: Greedy ::  Involving Sorting (Or The Input Is Already Sorted)
Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 1. Algorithm Design :: General Problem Solving Techniques ::  Examples
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Problem Solving Paradigms ::  Greedy - Standard
Root :: Prominent Problemsetters ::  Mohammad Mahmudur Rahman

简单模拟题。代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=1005;
struct JOB
{
    int b,j;
    bool operator <(const JOB &x) const
    {
        return j>x.j;//自定义<
    }
}job[maxn];
int main()
{
    int n,i,k,kase=0;
    while(scanf("%d",&n)!=EOF&&n)
    {
        for(i=0;i<n;i++)
            scanf("%d%d",&job[i].b,&job[i].j);
        sort(job,job+n);//默认<排序，所以重载函数故意这样写
        int ans=0,s=0;
        for(i=0;i<n;i++)
        {
             s+=job[i].b;
             ans=max(ans,s+job[i].j);//这一步的模拟是关键，有dp的感觉
        }
        printf("Case %d: %d\n",++kase,ans);
    }
    return 0;
}