HDOJ1789 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): 14870    Accepted Submission(s): 8679

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

 

Author

lcy

 

Source

2007省赛集训队练习赛（10）_以此感谢DOOMIII

 

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lcy   |   We have carefully selected several similar problems for you:  1257 1159 1069 1087 1231 

这题反正我想到的动归时间复杂度是N的3次方，那就算了.....

用贪心还是挺容易的。按照扣分从大到小排序，扣分一样当然是ddl小的在前面。

然后每次从一个作业的截止日期开始从后往前扫，如果有一天是空闲的话，意味着有时间完成作业，否则就要扣分。

#include <iostream>#include <cstdio>#include <cmath>#include <ctime>#include <algorithm>#include <cstring>using namespace std;const int maxn = 1005;const int inf = 1<<25;struct node {    int ddl,rs;}f[maxn];int n;int used[maxn],tot;bool cmp(node x, node y){    if (x.rs!=y.rs) return x.rs>y.rs;    else return x.ddl<y.ddl;}int main(){    int t,i,j,k;    std::ios::sync_with_stdio(false);    cin >> t;    while (t--) {        cin >> n;        for ( i=0; i<n; i++) cin >> f[i].ddl;        for ( i=0; i<n; i++) cin >> f[i].rs;        sort(f,f+n,cmp);        memset(used,0,sizeof(used));        tot = 0;        for ( i=0; i<n; i++) {                k = f[i].ddl;            for (j=k; j>0; j--) if (!used[j]) {                 used[j] =1;                 break;            }            if (j == 0) tot+=f[i].rs;        }        printf("%d\n",tot);    }    return 0;}