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

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

题目大意：n个部下，没人要完成一个任务。第i个任务要用bi时间交代，ji时间完成。要求交代不能同时，执行可以同时，求最小全部任务完成时间。

解题思路：执行时间较长的任务应该先执行，贪心，按j大从大到小对各个任务排序。

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<vector>
 4 using namespace std;
 5 struct Job{
 6     int j,b;
 7     bool operator<(const Job& x)const{
 8         return j>x.j;
 9     }
10 };
11 int main(){
12     int n,cases=1;
13     while(cin>>n && n){
14         vector<Job> V;
15         Job temp_job;
16         for(int i=0;i<n;i++){
17             cin>>temp_job.b>>temp_job.j;
18             V.push_back(temp_job);
19         }
20         sort(V.begin(),V.end());//按j从大到小排序
21         int s=0;
22         int ans=0;
23         for(int i=0;i<n;i++){
24             s+=V[i].b;//当前任务的开始时间
25             ans=max(ans,s+V[i].j);//执行完毕最晚时间
26         }
27         cout<<"Case "<<cases++<<": "<<ans<<'\n';            
28     }
29     return 0;
30 }

本文转自beautifulzzzz博客园博客，原文链接：http://www.cnblogs.com/zjutlitao/p/3595476.html，如需转载请自行联系原作者