poj 2029 Get Many Persimmon Trees(dp)
2017-08-09 11:09
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这道题由于数据量较小,全然能够暴力,用dp做先处理所有前缀,这样的复杂度是n*m,比暴力的复杂度低,暴力的复杂度是n*m*s*t
//#include<bits/stdc++.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
using namespace std;
int map1[105][105],dp[105][105];
int main()
{
int n;
while(~scanf("%d",&n)&&n)
{
int w,h,s,t,x,y;
scanf("%d%d",&w,&h);
memset(map1,0,sizeof(map1));
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
scanf("%d%d",&x,&y),map1[x][y]=1;
scanf("%d%d",&s,&t);
for(int i=1;i<=w;i++)
for(int j=1;j<=h;j++)
dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+map1[i][j];
int ans=0;
for(int i=1;i<=w-s+1;i++)
{
for(int j=1;j<=h-t+1;j++)
{
int ii=i+s-1;
int jj=j+t-1;
ans=max(ans,dp[ii][jj]-dp[ii][j-1]-dp[i-1][jj]+dp[i-1][j-1]);
}
}
printf("%d\n",ans);
}
return 0;
}
//#include<bits/stdc++.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
using namespace std;
int map1[105][105],dp[105][105];
int main()
{
int n;
while(~scanf("%d",&n)&&n)
{
int w,h,s,t,x,y;
scanf("%d%d",&w,&h);
memset(map1,0,sizeof(map1));
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
scanf("%d%d",&x,&y),map1[x][y]=1;
scanf("%d%d",&s,&t);
for(int i=1;i<=w;i++)
for(int j=1;j<=h;j++)
dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+map1[i][j];
int ans=0;
for(int i=1;i<=w-s+1;i++)
{
for(int j=1;j<=h-t+1;j++)
{
int ii=i+s-1;
int jj=j+t-1;
ans=max(ans,dp[ii][jj]-dp[ii][j-1]-dp[i-1][jj]+dp[i-1][j-1]);
}
}
printf("%d\n",ans);
}
return 0;
}
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