hdu_1789_Doing Homework again

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 
把扣分按照由大到小的顺序排列，之后从前到后扫描，放到相应天数的位置，如果那一天有作业了，那么就往它的前面的位置放，之后它没有位置可以放，说明这一天的作业无法完成。就把它加到计数器里面。
#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;struct node{    int x,y;};const int maxn=1002;node a[maxn];int vis[maxn]={0};bool cmp(node a,node b)//先按x从小到大排，相同，再按y从小到大{    return a.y>b.y||(a.y==b.y&&a.x<b.x);}int solve(int n){    int cnt=0,j;    for(int i=1;i<=n;i++)        {            for( j=a[i].x;j>0;j--) if(vis[j]==0)  {vis[j]=1;break;}            if(j<=0) cnt+=a[i].y;        }    return cnt;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        memset(vis,0,sizeof(vis));        int n;        scanf("%d",&n);        a[0].x=0;a[0].y=0;        for(int i=1;i<=n;i++) scanf("%d",&a[i].x);        for(int i=1;i<=n;i++) scanf("%d",&a[i].y);        sort(a+1,a+n+1,cmp);        printf("%d\n",solve(n));    }    return 0;}