UVA 11729 Commando War（求最少的时间）

11729 - Commando War

Time limit: 1.000 seconds

Thereis a war and it doesn't look very promising for your country. Now it's time toact. You have a commando squad at your disposal and planning an ambush on animportant enemy camp located nearby. You have
N soldiers in your squad.In your master-plan, every single soldier has a unique responsibility and youdon't want any of your soldier to know the plan for other soldiers so thateveryone can focus on his task only. In order to enforce this, you
brief everyindividual soldier about his tasks separately and just before sending him tothe battlefield. You know that every single soldier needs a certain amount oftime to execute his job. You also know very clearly how much time you need tobrief every single
soldier. Being anxious to finish the total operation as soonas possible, you need to find an order of briefing your soldiers that willminimize the time necessary for all the soldiers to complete their tasks. Youmay assume that, no soldier has a plan that depends
on the tasks of hisfellows. In other words, once a soldier begins a task, he can finish it without the necessity of pausing inbetween.

Input

Therewill be multiple test cases in the input file. Every test case starts with aninteger
N (1<=N<=1000), denoting the number of soldiers. Each ofthe following N lines describe a soldier with two integers
B(1<=B<=10000) & J (1<=J<=10000). B
secondsare 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 shouldnot be processed.

Output

Foreach test case, print a line in the format, “Case X: Y”, where X is the casenumber & Y is the total number of seconds counted from the start of yourfirst 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, SpecialThanks: Manzurur Rahman Khan

题意：求所有的任务完成的最少的时间（包括布置任务的时间）；

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
//author:YangSir
struct node{
int b,j;
}ca[1050];

bool cmp(node a,node b){
return a.j>b.j;
}
int max(int a,int b){
return a>b?a:b;
}
int main(){
int n,m,i,j,num=1,s,ans;
while(~scanf("%d",&n),n){//要时间最短，肯定要把完成任务时间最长的先布置完
for(i=0;i<n;i++)
scanf("%d%d",&ca[i].b,&ca[i].j);
sort(ca,ca+n,cmp);//按照完成任务的时间长短排序
s=0;
ans=0;
for(i=0;i<n;i++){
s+=ca[i].b;
ans=max(ans,s+ca[i].j);//不断比较出最短的时间
}
printf("Case %d: %d\n",num++,ans);
}
return 0;
}