Commando War ,UVa 11729

【题目】

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

【题解】

使用贪心算法，执行时间长的先交代，按照J从大到小的顺序给各个任务排序，然后依次交代。

#include <cstdio>
#include <vector>
#include <algorithm>

using namespace std;

struct Job {
int j, b;

bool operator<(const Job &x) const {             //运算符重载
return j > x.j;
}
};

int main() {
int n, b, j, kase = 1;
while (scanf("%d", &n) == 1 && n) {
vector<Job> v;
for (int i = 0; i < n; i++) {
scanf("%d%d", &b, &j);
v.push_back((Job) {j, b});
}
sort(v.begin(), v.end());           //使用Job类自己的<运算符排序
int s = 0;
int ans = 0;
for (int i = 0; i < n; i++) {
s += v[i].b;                //当前任务的开始执行时间
ans = max(ans, s + v[i].j);          //更新任务执行完毕时的最晚时间
}
printf("Case %d: %d\n", kase++, ans);
}

return 0;
}