hdu1789_Doing Homework again

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12185    Accepted Submission(s): 7162

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

333 3 310 5 131 3 16 2 371 4 6 4 2 4 33 2 1 7 6 5 4

 

Sample Output

035

 

题意：你有n门作业，每门作业需要一天的时间来做，并且这些作业有各自的截止日期 Di ，如果超期未交会扣 Ki 的分数，问你最少能扣多少分？

思路：贪心问题。先做分高的，在截止日期那一天做，如果那一天已经做了分更高的，就去从后向前面找没有做作业的一天，然后把作业做了。现在思考，是否可能我先安排一门分高的作业，导致后面安排的作业会超期。对于这个问题，我们可以想，没门作业都会做一天，所以如果前面安放的会让后面的放不下的话，那么也只会影响后面的一门，而如果我们把后面的课程先做完的话，就会导致前面分高的无法完成，就会扣除更多的分，所以上述的思路是可行的。但是这个题，比较麻烦的是他没有给课程天数的范围。。。有点尴尬，不过我试着开了一个1e7的数组，AC了，说明天数并没有超1e7，所以数据还是很水的。

#include<bits/stdc++.h>using namespace std;struct K{    int day;    int score;    int real;}s[1050];bool cmp(K a, K b){    return a.score > b.score;}int e[1000000];int main(){    int T;    scanf("%d",&T);    while(T--){        int n;        int sum = 0;        memset(e,0,sizeof(e));        scanf("%d",&n);        for(int i = 1; i <= n; i++)            scanf("%d",&s[i].day);        for(int i = 1; i <= n; i++)            scanf("%d",&s[i].score);        sort(s+1,s+1+n,cmp);        for(int i = 1; i <= n; i++){            int flag = 0;            for(int j = s[i].day; j >= 1; j--){                if(e[j] == 0){                    e[j] = 1;                    flag =1;                    break;                }            }            if(!flag)                sum += s[i].score;        }        cout << sum <<endl;    }}