HDU ACM 1789 Doing Homework again

原题描述

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

解题思路

将所有作业按照扣分数从大到小排序，先做扣分多的。

开一个数组记录当天是否被使用，然后枚举所有作业。

从一个作业的deadline向前枚举日期，如果发现空日期则做作业并占用日期，如果前面所有日期都满了则扣分。

累计扣分并输出。

参考代码

#include <iostream>
#include <algorithm>
using namespace std;

struct work             // 作业结构
{
int deadline;
int score;
bool operator< (const work & w)const    // 重载<，作为比较函数
{
// 扣分越高越靠前
return (score > w.score);
}
}W[1001];
int taken[1001];                        // 当天是否被使用

int main()
{
int t, n, i, j, ans;
cin >> t;
while(t--)
{
memset(taken, 0, sizeof(taken));// 0表示没有使用，1表示使用
ans = 0;
cin >> n;
for (int i = 0; i < n;i++)
{
cin >> W[i].deadline;
}
for (int i = 0; i < n;i++)
{
cin >> W[i].score;
}
sort(W, W + n);                 // 使用重载的<符号排序

for (i = 0; i < n;i++)      // 枚举每一个作业
{
for (j = W[i].deadline; j > 0;j--)      // 这个作业的deadline向前枚举日期，直到找到空的日子做作业
{
if(!taken[j])
{
taken[j] = 1;
break;
}
}
if (j == 0)                 // 如果没有找到空日子，那么扣分
ans += W[i].score;
}
cout << ans << endl;
}
}