HDU 1789 贪心

题目链接    ：http://acm.hdu.edu.cn/showproblem.php?pid=1789

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

 

就是m个课程，一行是课程结束时间，一行是课程的价值，求怎样没学课程价值最小

比赛时纠结好久误解，哎~还是太年轻了啊，感觉要多做些贪心的题目了

详解在代码中

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 1005struct stud{int time,cost;}f[N];int vis[N];int cmp(stud a,stud b){    if(a.cost!=b.cost)  //先做最大消耗的        return a.cost>b.cost;    return a.time<b.time;  //消耗相同就做时间靠前的}int main(){    int n,i,t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i=0;i<n;i++)            scanf("%d",&f[i].time);        for(i=0;i<n;i++)            scanf("%d",&f[i].cost);        sort(f,f+n,cmp);        memset(vis,0,sizeof(vis));          int j,ans=0;        for(i=0;i<n;i++)        {            j=f[i].time;            while(j)    //这里是重点，找到当前消耗的时间期限的前一个空时间，干掉这件事            {                if(vis[j]==0)                {                    vis[j]=1;                    break;                }                j--;            }            if(j==0)     //没有找到时间，这个任务无法完成               ans+=f[i].cost;        }        printf("%d\n",ans);    }    return 0;}