UVA 11729 Commando War

题目链接：Commando War

题面：

“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
3
2 5
3 2
2 1
3
3 3
4 4
5 5
0
Sample Output
Case 1: 8
Case 2: 15

题意：给士兵派发任务需要时间，完成时间，求完成总时间最短。

因为派发任务是连续的，所以派发的总时间是固定的，所以就是看执行时间。只要让执行时间长的先执行就好了。

代码：

#include <algorithm>
#include <iostream>
#include <cstdio>
using namespace std;
struct task
{
	int brief,times;
}store[10005];
bool cmp(task a,task b)
{
	return a.times>b.times;
}
int main()
{
	int n,maxx,p,tmp,cnt=1;
	while(scanf("%d",&n)&&n)
	{
	  maxx=p=tmp=0;
	  for(int i=0;i<n;i++)
	  {
  	     scanf("%d%d",&store[i].brief,&store[i].times);	
  	  }	
  	  sort(store,store+n,cmp);
  	  for(int i=0;i<n;i++)
  	  {
  	  	p+=store[i].brief;
  	  	tmp=p+store[i].times;
  	  	if(tmp>maxx)
  	  	maxx=tmp;
  	  }
  	  printf("Case %d: %d\n",cnt++,maxx);
	}
	return 0;
}