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Doing Homework again HDU 1789 (贪心)

2017-05-31 22:13 423 查看

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 93 Accepted Submission(s): 81

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

Author

lcy

题意：
有n门课，每一门课有一个限定的时间和超过时间后罚的分数，然后假定每一天只能完成一门课，问如何安排时间去完成这些课，使得罚分最少并输出。

为了让最后的结果最小，首先要把socre从大到小排序。然后每当考虑一门课，扫描期限内的天数（注意要从最后的期限往前枚举，给其它课尽可能的留下余地），如果能完成，就将该天标记，否则计入总罚分，以此类推。

#include <iostream>
#include <cmath>
#include <string>
#include <algorithm>
#include <string.h>
using namespace std;
struct Node {
int deadline;
int reduce;
}node[1005];
int visited[1005];
int main() {
int t;
int n;
cin >> t;
while (t--) {
int ans = 0;
memset(node, 0, sizeof(node));
memset(visited, 0, sizeof(visited));
cin >> n;
for (int i = 0; i < n; ++i) {
cin >> node[i].deadline;
}
for (int i = 0; i < n; ++i) {
cin >> node[i].reduce;
}

sort(node, node + n, [](const Node& n1,const Node& n2){return n1.reduce > n2.reduce || (n1.reduce == n2.reduce && n1.deadline < n2.deadline);});

for (int i = 0; i < n; ++i) {
int j = node[i].deadline;
for (; j >= 1; --j) {
if (!visited[j]) {
visited[j] = 1;
break;
}
}
if (j == 0)
ans += node[i].reduce;
}
cout << ans << endl;
}
}

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