hdu 1789 Doing Homework again（贪心）

Doing Homework again

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
3
3
3 3 3
10 5 1
3
1 3 1
6 2 3
7
1 4 6 4 2 4 3
3 2 1 7 6 5 4

Sample Output
0
3
5

思路1：
要使扣的分最少，那么应该按扣分从大到小排序，扣分相同则按截止日期从小到大排序
对每一门作业，如果在截止日期之前有空闲的天，则让其在这天完成这门作业即可，否则放到最后完成，并扣分

代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct node{    int deadline,reduce;}p[1010];int vis[1010];bool cmp(node a,node b){    if(a.reduce==b.reduce)        return a.deadline<b.deadline;    return a.reduce>b.reduce;}int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        memset(vis,0,sizeof(vis));        scanf("%d",&n);        for(int i=0;i<n;++i)            scanf("%d",&p[i].deadline);        for(int i=0;i<n;++i)            scanf("%d",&p[i].reduce);        sort(p,p+n,cmp);        int ans=0,j;        for(int i=0;i<n;++i)        {            for(j=p[i].deadline;j>0;--j)                if(!vis[j])            {                vis[j]=1;                break;            }            if(!j)                ans+=p[i].reduce;        }        printf("%d\n",ans);    }    return 0;}

思路2：
按照截止日期从小到大排序，截止日期相同则按扣分从大到小排序
对于每一门作业，如果当前日期大于截止日期，说明这门作业不能按时完成，这时可以在其截止日期之前寻找扣分较小的作业来代替它，这样使得最后的扣分最少

代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct node{    int x,y,flag;//flag代表这门作业是否能按时完成} p[1010];bool cmp(node a,node b){    if(a.x==b.x)        return a.y>b.y;    return a.x<b.x;}int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0; i<n; ++i)            scanf("%d",&p[i].x);        for(int i=0; i<n; ++i)            scanf("%d",&p[i].y),p[i].flag=1;        sort(p,p+n,cmp);        int reduced=0,day=1,tot,tmp;        for(int i=0; i<n; ++i)        {            if(day>p[i].x)            {                tot=p[i].y,tmp=i;                for(int j=0; p[j].x<=p[i].x&&j<i; ++j)                {                    if(p[j].y<tot&&p[j].flag)                    {                        tmp=j;                        tot=p[j].y;                    }                }                p[tmp].flag=0;                reduced+=tot;            }            else                ++day;        }        printf("%d\n",reduced);    }    return 0;}