HDU 1789 Doing Homework again（貪心演算法）

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4511    Accepted Submission(s): 2645


Problem DescriptionIgnatius 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. 

InputThe 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. 

OutputFor 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

 

Authorlcy 

Source2007省賽集訓隊練習賽（10）_以此感謝DOOMIII 

Recommendlcy 這個題目的貪心演算法比較簡單，也比較容易想到，就是按照扣除的分數從小到大排序然後從頭開始貪心的從每個位置的deadline向前尋找沒有被佔用的時間來做這門作業這樣貪心的保證時間盡量用靠後的，防止前面有的作業沒時間做，第二作業貪心的做扣分大的，這就是為什麼排序按照扣的分數從大到小的原因

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>using namespace std;struct point{    int re;    int deadline;}po[1500];bool rec[1500];int cmp(const void *a,const void *b){    return (*(point*)a).re < (*(point*)b).re ? 1 : -1;}int main(){    int t;    int n,i,j;    scanf("%d",&t);    int ans;    while(t--)    {        scanf("%d",&n);        for(i=0;i<n;i++)        scanf("%d",&po[i].deadline);        for(i=0;i<n;i++)        scanf("%d",&po[i].re);        qsort(po,n,sizeof(po[0]),cmp);        memset(rec,1,sizeof(rec));        ans=0;        for(i=0;i<n;i++)        {            for(j=po[i].deadline;j>=1;j--)            if(rec[j])            { rec[j]=false;break;}            if(j==0)            ans+=po[i].re;        }        printf("%d\n",ans);    }    return 0;}