HDU 1789(贪心)

问题描述:

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

题目题意:题目给我们n个作业期限时间和它不完成的代价，每一天可以完成一个作业，作业不完成就会得到它的代价，问最小的代价是多少？

题目分析:很明显，代价越高的作业应该优先完成，代价相等的应该先完成期限短的，我们按照要求排序后，遍历每个作业，时间上从它的期限往前遍历，找到了一个没用过的时间就可以做这次作业了，反之加上代价。

代码如下:

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int maxn=1e3+100;struct note{    int t,v;}point[maxn];bool vis[maxn];bool cmp(struct note a,struct note b){    if (a.v==b.v) return a.t<b.t;    else return a.v>b.v;}int main(){    int t,n;    scanf("%d",&t);    while (t--) {        scanf("%d",&n);        for (int i=0;i<n;i++) {            scanf("%d",&point[i].t);        }        for (int i=0;i<n;i++) {            scanf("%d",&point[i].v);        }        sort(point,point+n,cmp);        memset (vis,false,sizeof (vis));        int ans=0;        for (int i=0;i<n;i++) {            bool flag=false;            for (int j=point[i].t;j>=1;j--) {                if (!vis[j]) {                    vis[j]=true;                    flag=true;                    break;                }            }            if (!flag) ans+=point[i].v;        }        printf("%d\n",ans);    }    return 0;}