不充钱,你怎么AC?
题目:http://codevs.cn/problem/2456/
用贪心的思想,木材当然要尽量分成多的木板,而大的木材能够分成大木板,但是小的木材不一定能够分成大的木板,所以木板和木材都是从小到大开始选,然后要保证剩余的木材最少
那么将木板和木材排序,对于每个木材,把能够分的小木板尽量分掉,如果遇到更大的木板则把最小的木板腾出来,然后在加上,这样保证剩余的木材最少
因为是上午写得这道题,思路可能不连贯了,代码应该描述的很清楚
虽然贪心在CodeVS上可以过,但是它是有数据可以卡掉的,而正解是DFS
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std; const int N=,M=;
int a[N],b[M],next[M];
bool f[M];
int main()
{
int n,i,j,m,k,now,ans=,x,y;
scanf("%d",&n);
for (i=;i<=n;i++) scanf("%d",&a[i]);
sort(a+,a++n);
scanf("%d",&m);
for (i=;i<=m;i++) scanf("%d",&b[i]);
sort(b+,b++m);
for (i=;i<=n;i++)
{
now=a[i];
for (j=;j<=m;j++) next[j]=;
for (j=;j<=m;j++)
{
if (f[j]) continue;
if (now>=b[j])
{
now-=b[j];
f[j]=;
ans++;
next[j]=next[];
next[]=j;
}
else
{
k=x=;
while (now+b[next[k]]>=b[j]) k=next[x=k];
if (k)
{
next[x]=next[k];
f[k]=;
f[j]=;
next[j]=next[];
next[]=j;
now+=b[k]-b[j];
}
}
}
}
printf("%d\n",ans);
return ;
}
下面是DFS的正解,贴的别人的代码
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std; #define N 1100 int n,m;
int sum1,ans;
int tim,res; int a[N],b[N],c[N],sum[N]; int work(int x,int k)
{
if (!x)
return true;
if (tim>res)
return false;
for (int i=k;i<=n;i++)
if (a[i]>=b[x])
{
a[i]-=b[x];
if (a[i]<b[])
tim+=a[i];
if (b[x]==b[x-])
{
if (work(x-,i))
return true;
}
else if (work(x-,))
return true;
if (a[i]<b[])
tim-=a[i];
a[i]+=b[x];
}
return false;
} int main()
{
scanf("%d",&n);
for (int i=;i<=n;i++)
scanf("%d",&a[i]),sum1+=a[i],c[i]=a[i];
scanf("%d",&m);
for (int i=;i<=m;i++)
scanf("%d",&b[i]);
sort(a+,a+n+);
sort(b+,b+m+);
for (int i=;i<=m;i++)
sum[i]=b[i]+sum[i-];
int l=,r=m;
while (l!=r)
{
int mid=(l+r+)>>;
tim=;
res=sum1-sum[mid];
if (work(mid,))
l=mid;
else
r=mid-;
for (int i=;i<=n;i++)
a[i]=c[i];
}
printf("%d\n",l);
return ;
}