HDU 1789 Doing Homework again (贪心)
Doing Homework again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6499 Accepted Submission(s): 3874
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.
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
Author
lcy
Source
Recommend
lcy
这道题的贪心策略是先取减分多的，然后取减分少的
所以先要排序，把减分多的排到前面，如果减分相同，把deadline小的放前面。
然后我每次取出一个任务，如果能在deadline那一天执行，就放在那一天执行，不行的话，就往前找，找到能执行的那一天就在那一天执行。如果所有能执行的天数都有其他的任务的话，那么我这个任务就不能完成，就要减去相应的分数。
1 #include<cstdio> 2 #include<cstring> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 struct node 7 { 8 int day; 9 int num; 10 bool operator<(const node &B)const 11 { 12 if(num!=B.num) 13 return num>B.num; 14 else 15 return day<B.day; 16 } 17 }a[1000]; 18 int vis[1000]; 19 int main() 20 { 21 int kase; 22 scanf("%d",&kase); 23 while(kase--) 24 { 25 int n,i,j,sum=0,ans=0; 26 memset(vis,0,sizeof(vis)); 27 scanf("%d",&n); 28 for(i=1;i<=n;i++) 29 scanf("%d",&a[i].day); 30 for(i=1;i<=n;i++) 31 scanf("%d",&a[i].num); 32 sort(a+1,a+n+1); 33 for(i=1;i<=n;i++) 34 { 35 for(j=a[i].day;j>0;j--) 36 { 37 if(!vis[j]) 38 { 39 vis[j]=1; 40 break; 41 } 42 } 43 if(j==0) 44 ans+=a[i].num; 45 } 46 printf("%d\n",ans); 47 } 48 return 0; 49 }
