这题是一道比较简单的动态规划,类似的可以有最大子矩形和。
与另外一道triangles也比较像。
#include <iostream>
#define N 201
using namespace std;
char gh[N][N];
int row,col,n;
int rf[N][N][2];
void input()
{
int i,j;
for (i=0;i<n;i++)
{
for (j=0;j<2*n-i*2-1;j++)
{
cin>>gh[i][j];
}
}
for (i=0;i<n;i++)
{
for (j=0;j<col;j++)
{
rf[i][j][0]=rf[i][j][1]=-1;
}
}
}
int f (int i,int j,int k)
{
if (i<0||i>=n||j<0||j>=col-i*2)return 0;
if (i==0)return gh[i][j]=='-';
if (gh[i][j]=='#')return 0;
if (rf[i][j][k]!=-1)return rf[i][j][k];
if (gh[i-1][j+1]=='#')
{
rf[i][j][k]=1;
}
else
{
rf[i][j][k]=1+min(f(i-1,j,k),f(i-1,j+2,k));
}
return rf[i][j][k];
}
int g(int i,int j,int k)
{
if (i<0||i>=n||j<0||j>=col-i*2)return 0;
if (i==n-2)return gh[i][j]=='-';
if (gh[i][j]=='#')return 0;
if (rf[i][j][k]!=-1)return rf[i][j][k];
if (gh[i+1][j-1]=='#')
{
rf[i][j][k]=1;
}
else
{
rf[i][j][k]=1+min(g(i+1,j,k),g(i+1,j-2,k));
}
return rf[i][j][k];
}
int dp()
{
int mV=0,t,i,j;
for (i=0;i<n;i++)
{
for (j=0;j<col-2*i;j+=2)
{
if ((t=f(i,j,0))>mV)
mV=t;
}
for (j=1;j<col-2*i;j+=2)
{
if ((t=g(i,j,1))>mV)
mV=t;
}
}
return mV;
}
int main()
{
int result;
int ncase=0;
while (cin>>n)
{
ncase++;
if (n==0)break;
row=n;
col=2*n-1;
input();
result=dp();
cout<<"Triangle #"<<ncase<<endl;
cout<<"The largest triangle area is "<<result*result<<"."<<endl<<endl;
}
return 0;
}
#define N 201
using namespace std;
char gh[N][N];
int row,col,n;
int rf[N][N][2];
void input()
{
int i,j;
for (i=0;i<n;i++)
{
for (j=0;j<2*n-i*2-1;j++)
{
cin>>gh[i][j];
}
}
for (i=0;i<n;i++)
{
for (j=0;j<col;j++)
{
rf[i][j][0]=rf[i][j][1]=-1;
}
}
}
int f (int i,int j,int k)
{
if (i<0||i>=n||j<0||j>=col-i*2)return 0;
if (i==0)return gh[i][j]=='-';
if (gh[i][j]=='#')return 0;
if (rf[i][j][k]!=-1)return rf[i][j][k];
if (gh[i-1][j+1]=='#')
{
rf[i][j][k]=1;
}
else
{
rf[i][j][k]=1+min(f(i-1,j,k),f(i-1,j+2,k));
}
return rf[i][j][k];
}
int g(int i,int j,int k)
{
if (i<0||i>=n||j<0||j>=col-i*2)return 0;
if (i==n-2)return gh[i][j]=='-';
if (gh[i][j]=='#')return 0;
if (rf[i][j][k]!=-1)return rf[i][j][k];
if (gh[i+1][j-1]=='#')
{
rf[i][j][k]=1;
}
else
{
rf[i][j][k]=1+min(g(i+1,j,k),g(i+1,j-2,k));
}
return rf[i][j][k];
}
int dp()
{
int mV=0,t,i,j;
for (i=0;i<n;i++)
{
for (j=0;j<col-2*i;j+=2)
{
if ((t=f(i,j,0))>mV)
mV=t;
}
for (j=1;j<col-2*i;j+=2)
{
if ((t=g(i,j,1))>mV)
mV=t;
}
}
return mV;
}
int main()
{
int result;
int ncase=0;
while (cin>>n)
{
ncase++;
if (n==0)break;
row=n;
col=2*n-1;
input();
result=dp();
cout<<"Triangle #"<<ncase<<endl;
cout<<"The largest triangle area is "<<result*result<<"."<<endl<<endl;
}
return 0;
}