HDOJ-----1789贪心

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11320    Accepted Submission(s): 6649

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

03
5

就是老师布置作业，规定时间前没做完就扣分，一门作业要做一天，问最少扣多少分

贪心的变形

先按规定时间升序排列，时间一样，再按扣分降序排列

就好比高数作业星期一截止，没做完扣一分，英语作业星期二截止，没做完扣20分。物理作业星期三截止没做完扣90分，

语文作业也是星期三截止，没做完扣60分，就要用星期一做语文作业，星期二做英语作业，星期三做物理作业，高数老师

罚就让他罚吧，反正扣一分

嗯，对高数满满的恶意--------

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct node{    int a, b, c;}s[10010];bool cmp(node x, node y){    if(x.a == y.a){        return x.b > y.b;    }    return x.a < y.a;}int main(){    int n, m;    scanf("%d", &n);    while(n--){        memset(s, 0, sizeof(s));        scanf("%d", &m);        for(int i = 0; i < m; i++){            scanf("%d", &s[i].a);        }        for(int i = 0; i < m; i++){            scanf("%d", &s[i].b);            s[i].c = 1;        }        sort(s, s + m, cmp);        int k = 1, ok = 1, ans = 0, j, t;        for(int i = 0; i < m; i++){            if(s[i].a >= k){//先把第k天扣分最多的做完                k++;                continue;            }            ok = s[i].b;            t = i;            for(j = 0; j < i; j++){                if(ok > s[j].b && s[j].c){//如上所述，取出前几天扣分最少的加入扣的总分里，并标记分已被扣除，下次要用扣分倒数第二低的来补                    ok = s[j].b;                    t = j;                }            }            ans += ok;            s[t].c = 0;        }        printf("%d\n", ans);    }    return 0;}