UVA_11729Commando War

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

[align=center][/align]

思路：贪心的策略，让执行时间长的任务先做。例如X和Y两个任务，前提是J[X] > J[Y] ,分别证明:1.当B[X] > B[Y] ,2.B[X] == B[Y],3.B[X] < B[Y]是分别交换X和Y的交待顺序，最后发现只要X的交待顺序在Y之前完成两个任务的总时间总是最短的。

#include<cstdio>
#include<algorithm>
using namespace std;

struct JOB{//定义一个Job数组用来表示表示每个任务所需要的吩咐时间和执行时间，同时再加一个运算符重载
int b, j;
bool operator < (const JOB & a) const{
return j > a.j;
}
}Job[1005];

int main()
{
int N, B, J;
int count = 1;
while(scanf("%d", &N) == 1 && N){
for(int i = 0; i < N; i++){
scanf("%d%d", &B, &J);//将每个任务的数据输入
Job[i] = JOB{B, J};
}
sort(Job, Job+N);
int s = 0;
int ans = 0;
for(int i = 0; i < N; i++){
s += Job[i].b;
ans = max(ans, s+Job[i].j);
}
printf("Case %d: %d\n", count++, ans);
}
return 0;
}