Doing Homework again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6577 Accepted Submission(s):
3926
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
思路：按照贪心的思想，每次做出最优决策，所以可以将作业按扣分高低排序，每次找到扣分最高的，将它安排进终止日期那一天，如果那天有事，就在往前推，，如果最后安排不下，就扣分。这样循环n次就可以了。
1 #include <cstdio> 2 #include <string.h> 3 #include <algorithm> 4 using namespace std; 5 int T, N, hash[1010]; 6 struct HM 7 { 8 int day, score; 9 } hm[1010]; 10 bool cmp (HM x, HM y ) 11 { 12 if ( x.score != y.score ) 13 return x.score > y.score; 14 return x.day > y.day; 15 } 16 int main () 17 { 18 scanf ( "%d", &T ); 19 int i,j; 20 while ( T -- ) 21 { 22 scanf ( "%d", &N ); 23 int sum = 0; 24 for (i=1; i <= N; ++ i ) 25 scanf ( "%d", &hm[i].day ); 26 for (i=1; i <= N; ++ i ) 27 scanf ( "%d", &hm[i].score ); 28 sort ( hm+1, hm+N+1, cmp ); 29 memset ( hash, 0, sizeof ( hash ) ); 30 for (i=1; i<= N; ++i ) 31 { 32 for ( j = hm[i].day; j > 0; j-- ) 33 { 34 if ( !hash[j] ) 35 { 36 hash[j]=1; 37 break; 38 } 39 } 40 if ( !j ) 41 sum += hm[i].score; 42 } 43 printf ( "%d ", sum ); 44 } 45 return 0; 46 }