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Uva 11729 Commando War【贪心】

来源：程序员人生发布时间：2016-10-12 09:19:03 阅读次数：2202次

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

2 5

3 2

2 1

3 3

4 4

5 5

Case 1: 8

Case 2: 15

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

题意：有n的任务，每一个任务安排需要b秒，完成任务需要j秒，不能同时安排任务，但能同时做不同的任务，问你做完所有任务的最少时间。

贪心，根据完成任务时间从大到小排序，1次进行便可。

对数据的贮存，可以用结构体数组，也能够用vector(白书上例题)。

Vector AC代码：

#include <iostream> #include <cstdio> #include <algorithm> #include <vector> using namespace std; struct Node { int s,l; Node(int s=0,int l=0):s(s),l(l){} bool operator <(const Node& x) const { return l>x.l; } }; int main() { int kase=0; int n; ios::sync_with_stdio(false); cin.tie(0); while(cin>>n,n) { vector<Node>v; int x,y; for(int i=0;i<n;i++) { cin>>x>>y; v.push_back(Node(x,y)); } sort(v.begin(),v.end()); int ans=0; int start=0; for(int i=0;i<n;i++) { start+=v[i].s; ans=max(ans,start+v[i].l); } printf("Case %d: %d\n",++kase,ans); } return 0; }

数组 AC代码：

#include <iostream> #include <cstdio> #include <algorithm> using namespace std; struct node { int s,l; }a[1005]; bool cmp(node x,node y) { return x.l>y.l; } int main() { int kase=0; int n; ios::sync_with_stdio(false); cin.tie(0); while(cin>>n,n) { int x,y; for(int i=0;i<n;i++) { cin>>a[i].s>>a[i].l; } sort(a,a+n,cmp); int ans=0; int start=0; for(int i=0;i<n;i++) { start+=a[i].s; ans=max(ans,start+a[i].l); } printf("Case %d: %d\n",++kase,ans); } return 0;

尊重原创，转载请注明出处：http://blog.csdn.net/hurmishine

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