期末考试(优先队列)

期末考试

时间限制：1000 ms  |  内存限制：65535 KB

难度：2

描述

马上就要考试了，小T有许多作业要做，而且每个老师都给出来了作业要交的期限，如果在规定的期限内没

交作业就会扣期末成绩的分数，假设完成每门功课需要一天的时间，你能帮助小T扣除的分数最小吗？

输入

输入n，表示n门功课(n<2000)，接下来n行，每行两个数a，b，分别表示交作业的最后期限，迟交扣除的分数。

(以文件结尾)

输出

输出扣除的最小分数。

样例输入

3
3 10
3 5
3 1
3
1 6
3 2
1 3
7
1 3
4 2
6 1
4 7
2 6
4 5
3 4      

样例输出

0
3
5      

题解：sort先排序，然后判断扣分值，先做分高的；优先队列从小到大，长度代表天；

代码：

1 #include<stdio.h>
 2 #include<queue>
 3 #include<algorithm>
 4 #include<string.h>
 5 using namespace std;
 6 struct Node{
 7     int time,score;
 8 };
 9 int cmp(Node a,Node b){
10     if(a.time!=b.time)return a.time<b.time;
11     else return a.score>b.score;
12 }
13 Node m[2010];
14 int main(){
15     int n,day,tot;
16     while(scanf("%d",&n),n){priority_queue<int,vector<int>,greater<int> >work;tot=0;
17     memset(m,0,sizeof(m));
18         for(int i=0;i<n;++i)scanf("%d%d",&m[i].time,&m[i].score);
19         sort(m,m+n,cmp);
20         for(int i=0;i<n;++i){day=work.size();
21             if(m[i].time>day)work.push(m[i].score);
22             else{
23                 if(!work.empty()&&m[i].score>work.top())tot+=work.top(),work.pop(),work.push(m[i].score);
24             else tot+=m[i].score;
25             }
26         }
27         printf("%d\n",tot);
28     }
29     return 0;
30 }      

Doing Homework again

Time Limit : 1000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 66   Accepted Submission(s) : 44

Problem Description

Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test. And now we assume that doing everyone homework always takes one day. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.

Input

The input contains several test cases. The first line of the input is a single integer T that is the number of test cases. T test cases follow. Each test case start with a positive integer N(1<=N<=1000) which indicate the number of homework.. Then 2 lines follow. The first line contains N integers that indicate the deadlines of the subjects, and the next line contains N integers that indicate the reduced scores.

Output

For each test case, you should output the smallest total reduced score, one line per test case.

Sample Input

3 3 3 3 3 10 5 1 3 1 3 1 6 2 3 7 1 4 6 4 2 4 3 3 2 1 7 6 5 4

Sample Output

0 3 5

 题解，两道题差不多：

1 #include<stdio.h>
 2 #include<math.h>
 3 #include<algorithm>
 4 using namespace std;
 5 struct Node{
 6     double s,e;
 7 };
 8 int n,d,k;
 9 Node area[1010];
10 int cmp(Node a,Node b){
11     return a.e<b.e;
12 }
13 int change(int x,int y){
14     if(y>d)return 0;
15     double a,b,m=sqrt(d*d-y*y);
16     a=x-m;b=x+m;
17     area[k].s=a;area[k].e=b;k++;
18     return 1;
19 }
20 int main(){int t,x,y,flot,temp,num,l=0;
21     while(scanf("%d%d",&n,&d),n||d){k=0;flot=1;temp=0;num=1;l++;
22             for(int i=0;i<n;i++){
23                 scanf("%d%d",&x,&y);
24                 t=change(x,y);
25                 if(!t)flot=0;
26             }
27             sort(area,area+k,cmp);
28             for(int i=0;i<k;i++){
29                 if(area[i].s>area[temp].e)temp=i,num++;
30             }
31             printf("Case %d: %d\n",l,num);
32     }
33     return 0;
34 }