hdu1789 sdut2076 Doing Homework again（贪心）

Doing Homework again

Time Limit: 1000MS    Memory limit: 65536K

题目描述

 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.

输入

 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.

输出

 For each test case, you should output the smallest total reduced score, one line per test case.

示例输入

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

示例输出

0
3
5

 

题意：

 一天只能完成一个作业，每个作业有规定的完成时间和未完成要扣的分数，求扣分最小值

 

方法一：

先按时间排，时间相同的分高优先，剩余完不成的与以前完成的作业分数比较，如果分数较大则替换

#include<stdio.h>
#include<algorithm>
using namespace std;
struct ss
{
int dead,sco,ok;
} work[1010];
int cmp(ss a,ss b)
{
if(a.dead==b.dead) return a.sco>b.sco;
return a.dead<b.dead;
}
int main()
{
int t,n,i,j,sum,day,a,b;
while(scanf("%d",&t)!=EOF)
{
while(t--)
{
scanf("%d",&n);
for(i=0; i<n; i++)
{
scanf("%d",&work[i].dead);
work[i].ok=0;
}
for(i=0; i<n; i++)
{
scanf("%d",&work[i].sco);
}
sort(work,work+n,cmp);
day=1;
for(i=0; i<n; i++)
{
if(work[i].dead>=day)
{
work[i].ok=1;
day++;
}
}
for(i=0; i<n; i++)
{
if(work[i].ok==0)
{
a=i;
b=work[i].sco;
for(j=0; j<i; j++)
{
if(work[j].ok==1&&work[j].sco<b)
{
a=j;
b=work[j].sco;
}
}
if(a!=i)
{
work[i].ok=1;
work[a].ok=0;
}
}
}
sum=0;
for(i=0; i<n; i++)
{
if(work[i].ok==0)
{
sum+=work[i].sco;
}
}
printf("%d\n",sum);
}
}
return 0;
}

 方法二：

从最后期限那天开始选择作业，一直选到第一天，手工模拟一下很容易理解

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct ss
{
int dea,sco;
bool don;
} s[1004];
bool cmp(ss a,ss b)
{
if(a.dea==b.dea) return a.sco>b.sco;
return a.dea>b.dea;
}
int main()
{
int t,n,i,j,x,ans,mx;
while(scanf("%d",&t)!=EOF)
{
while(t--)
{
memset(s,0,sizeof(s));
scanf("%d",&n);
for(i=0; i<n; i++)
scanf("%d",&s[i].dea);
for(i=0; i<n; i++)
scanf("%d",&s[i].sco);
sort(s,s+n,cmp);
for(i=s[0].dea; i>0; i--)//假如时光能倒流，我们从最后期限开始选择作业
{
for(mx=x=j=0; j<n; j++)
{
if(s[j].dea<i)
break;
if(!s[j].don&&s[j].sco>mx)
{
mx=s[j].sco;
x=j;
}
}
s[x].don=1;
}
for(ans=i=0; i<n; i++)
if(!s[i].don)
ans+=s[i].sco;
printf("%d\n",ans);
}
}
return 0;
}