Commando War

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题目链接: 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     这是一道贪心题目,大意就是每个士兵有B秒的准备时间和J秒的工作时间,准备时间不能重叠,给出几个士兵的准备时间和工作时间,求从开始到完成工作需要多长时间。 贪心思路:对J从大到小排序。第一个士兵完成工作所用时间为B1+J1;第二个士兵:B1+B2+J2。第三个。。。,从中找出最长时间即为所求结果。 #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; struct sold { int b,j; }a[1100]; bool cmp1(sold x,sold y) { return (x.j > y.j); } int main() { int i,n,m,x,y,b[10000],t,f,sum,ch=1; while(scanf("%d",&n)!=EOF) { if(n==0) break; else { sum=0; memset(b,0,sizeof(b)); for(i=0;i<n;i++) scanf("%d %d",&a[i].b,&a[i].j); sort(a,a+n,cmp1); for(t=0;t<n;t++) { sum=0; for(f=t;f>=0;f--) sum=sum+a[f].b; b[t]=sum+a[t].j; } printf("Case %d: ",ch); ch++; //printf("%d\n",b[n-1]); sort(b,b+t); printf("%d\n",b[n-1]); } } return 0; }    



关键词: Commando War