HDU1789 贪心
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Doing Homework again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14162 Accepted Submission(s): 8216
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
333 3 310 5 131 3 16 2 371 4 6 4 2 4 33 2 1 7 6 5 4
Sample Output
035
以作业的分值为第一关键字,作业的最后期限为第二关键字排序。从后往前安排时间,将时间期限用一个数组来标记,如果遍历到最前面都没有把作业安排进去,就说明作这份作业不划算。
#include <iostream>#include <algorithm>#include <cstring>using namespace std;typedef pair<int, int> P;bool cmp(const P a, const P b){ if(a.second == b.second) return a.first < b.first; return a.second > b.second;}int main(){ cin.tie(0); cout.tie(0); P p[1007]; bool flg[1007]; int T; cin >> T; while(T--){ memset(flg, 0, sizeof(flg)); int n; cin >> n; for(int i = 0; i < n; i++) cin >> p[i].first; for(int i = 0; i < n; i++) cin >> p[i].second; sort(p, p + n, cmp); int ans = 0; for(int i = 0; i < n; i++){ int j; for(j = p[i].first - 1; j >= 0; j--) if(!flg[j]){ flg[j] = true; break; } if(j == -1) ans += p[i].second; } cout << ans << endl; }}
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