HDUoj 1789 Doing Homework again ( 贪心好题

2019-11-11 07:08:10

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Doing Homework again

Time Limit: 1000/1000 MS (java/Others)

Memory Limit: 32768/32768 K (Java/Others)

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. 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
 0 3 5 怎么才能提高一个智障的人的智商啊
/*    先按照分数从大到小排序，分数相等就按照天数从小到大排序*/ #include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define N 1000000bool arr[N];       // 标记数组，两个值就够了 struct node{    int a, b;}p[1000];bool cmp(node x, node y){    if(x.b != y.b)        return x.b > y.b;         //分数不相等，从大到小排序 return x.a < y.a;}int main(){    int T, n;    scanf("%d",&T);    while(T--){        bool flag;        memset(arr,false,sizeof(arr));        scanf("%d",&n);        for(int i = 0;i < n; i++)            scanf("%d",&p[i].a);        for(int i = 0;i < n; i++)            scanf("%d",&p[i].b);        sort(p,p+n,cmp);        int k = 0;        for(int i = 0;i < n; i++){            flag = false;            for(int j = p[i].a;j > 0; j--){                if(!arr[j]){                    arr[j] = true;                    flag = true;                    break;                }            }            if(!flag)            k += p[i].b;        }        printf("%d/n",k);    }return 0;}

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