HDU 1789 Doing Homework again(贪心)

Doing Homework again

Time Limit : 1000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 5   Accepted Submission(s) : 3

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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

Author

lcy

Source

2007省赛集训队练习赛（10）_以此感谢DOOMIII

#include<stdio.h>  #include<string.h>  #include<algorithm>  using namespace std;  struct work  {      int time,score;  } a[5001];  //先按分数从大到小排序，分数相同下按时间从小到大 bool cmp(work a,work b)  {      if(a.score==b.score)          return a.time<b.time;      else          return a.score>b.score;  }  int main()  {      int visit[1001]={0};  //标志     int n,m,i,s,j;      scanf("%d",&n);      while(n--)      {          scanf("%d",&m);                  memset(visit,0,sizeof(visit));    //标志初始位                 for(i=0; i<m; i++)       //输入时间             scanf("%d",&a[i].time);                       for(i=0; i<m; i++)     //输入扣分数             scanf("%d",&a[i].score);                      sort(a,a+m,cmp);   //排序                 //先选扣分数多的，依次标记所用的日期         s=0;          for(i=0;i<m;i++)          {              j=a[i].time;                 while(j)              {                  if(visit[j]==0)                  {                      visit[j]=1;                      break;                  }                  j--;              }              if(j==0)                  s+=a[i].score;          }          printf("%d\n",s);      }  } 例子：分数：7 6 5 4 3 2 1时间：4 2 4 2 1 4 6j = 4   visit[4] = 1;j = 2   visit[2] = 1;j = 4   visit[3] = 1;j = 2visit[1] = 1;j = 1   j--;j = 0    s+=3j = 4   j--;j = 0    s+=2j = 6   visit[6] = 1;

附上今天错误代码

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>#include<cstdlib>using namespace std;struct node{    int t,w,vis;}a[1005];int cmp(node a,node b){    if(a.t>b.t)    return 1;    else if (a.t==b.t&&a.w>b.w)    return 1;    return 0;}int main(){    //freopen("in.txt","r",stdin);int t,n,i,j;scanf("%d",&t);while(t--){       int maxtime,sum=0;    memset(a,0,sizeof(a));    scanf("%d",&n);        for(int i=0;i<n;i++)    {    scanf("%d",&a[i].t);    maxtime=max(maxtime,a[i].t);    }        for(int i=0;i<n;i++)    {    scanf("%d",&a[i].w);    sum+=a[i].w;    }    sort(a,a+n,cmp);        int ans=0;    for(i=maxtime;i>0;i--)    {        int maxw=0,flag=0;        for(j=0;j<n;j++)        {            if(a[j].t>=i&&a[j].vis==0)            {                if(maxw<a[j].w)                {                maxw=a[j].w;                flag=j;                }                //printf("%d\n",ans);            }        }        if(a[flag].vis==0)        {            ans+=a[flag].w;            //printf("%d\n",ans);            a[flag].vis=1;        }    }    printf("%d\n",sum-ans);}return 0;}