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): 4039    Accepted Submission(s): 2335

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

 

简单贪心题目，思路是先按期末减少分数从大到小排序，开辟一个数组cnt用于标记。然后从第零个开始处理，将cnt[第零个的期限]标记，然后每一次都判断一下cnt[第i个期限]是否为零，如果不是则判断cnt[第i个期限-1]，如果所有的数组都不为零。说明一定要放弃它了，让sum+它一下。以此类推即可。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MN=1010;
struct home
{
    int deadline;
    int reduce;
}node[1010];
int cnt[MN];
bool cmp(struct home a,struct home b)
{
    if(a.reduce==b.reduce)
        return a.deadline<b.deadline;
    else return a.reduce>b.reduce;
}
int main()
{
    int cas,n,flag,sum;
    scanf("%d",&cas);
    while(cas--){
        memset(cnt,0,sizeof(cnt));
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&node[i].deadline);
        for(int i=0;i<n;i++)
            scanf("%d",&node[i].reduce);
        sort(node,node+n,cmp);
        sum=0;
        for(int i=0;i<n;i++){
            flag=0;
            if(!cnt[node[i].deadline])
                cnt[node[i].deadline]=1;
            else{
                for(int j=node[i].deadline-1;j>=1;j--){
                    if(!cnt[j]){
                        cnt[j]=1;flag=1;break;
                    }
                }
                if(!flag) sum+=node[i].reduce;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}