usaco Home on the Range
2016-12-04 22:34
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dp关系式搞清楚就行了,我看题解的没想出来。a[i][j]=min(a[i+1][j],a[i][j+1],a[i+1][j+1])+1;他的右方下方右下方保证了他可以延伸的长度。/*
ID: jinbo wu
TASK: range
LANG:C++
*/
#include<bits/stdc++.h>
using namespace std;
int a[300][300];
int cnt[300];
int main()
{
freopen("range.in","r",stdin);
freopen("range.out","w",stdout);
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
scanf("%1d",&a[i][j]);
}
for(int i=n-1;i>=1;i--)
for(int j=n-1;j>=1;j--)
{
if(a[i][j])
a[i][j]=min(a[i+1][j],min(a[i][j+1],a[i+1][j+1]))+1;
}
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
while(a[i][j]>=2)//大正方形包括了小正方形
{
cnt[a[i][j]]++;
a[i][j]--;
}
}
for(int i=2;i<=n;i++)
{
if(cnt[i])
{
printf("%d %d\n",i,cnt[i]);
}
}
}
ID: jinbo wu
TASK: range
LANG:C++
*/
#include<bits/stdc++.h>
using namespace std;
int a[300][300];
int cnt[300];
int main()
{
freopen("range.in","r",stdin);
freopen("range.out","w",stdout);
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
scanf("%1d",&a[i][j]);
}
for(int i=n-1;i>=1;i--)
for(int j=n-1;j>=1;j--)
{
if(a[i][j])
a[i][j]=min(a[i+1][j],min(a[i][j+1],a[i+1][j+1]))+1;
}
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
while(a[i][j]>=2)//大正方形包括了小正方形
{
cnt[a[i][j]]++;
a[i][j]--;
}
}
for(int i=2;i<=n;i++)
{
if(cnt[i])
{
printf("%d %d\n",i,cnt[i]);
}
}
}
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