hdu 4328 cut the cake#单调队列#DP

/**
找最大的纯0 纯1 或01交叉的矩阵，01交叉的要是方阵

前两种单调栈或单调队列
第三种dp
dp[i][j] = min(dp[i][j-1],dp[i-1][j]) + (mat[i][j] == mat[i-k][j-k])
*/
#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
#define N 1100

char s

;
int mat

;
int lef
,rig
,dp

;
int m,n,hg[2]
;

int f(int flag)
{
memset(hg,0,sizeof(hg));
int j;
int cur,pre,tmp,res = 0;
for(int i = 1; i <= n; ++i)
{
cur = (i & 1);
pre = !cur;
for(j = 1; j <= m; ++j)
if(mat[i][j] == flag)
hg[cur][j] = hg[pre][j] + 1;
else
hg[cur][j] = 0;

for(j = 1; j <= m; ++j)
lef[j] = rig[j] = j;
for(j = 2; j <= m; ++j)
while(lef[j] > 1 && hg[cur][j] <= hg[cur][lef[j]-1])
lef[j] = lef[lef[j]-1];
for(j = m - 1; j; --j)
while(rig[j] < m && hg[cur][j] <= hg[cur][rig[j]+1])
rig[j] = rig[rig[j]+1];
for(j = 1; j <= m; ++j)
{
if(hg[cur][j] == 0)///这让我想说卧槽~~~，疏忽鸟，作死WA
continue;
tmp = (rig[j] - lef[j] + 1 + hg[cur][j]) * 2;
if(tmp > res)
res = tmp;
}
}
return res ;
}

int dd()
{
int i,j,ans = 1;
for(i = 1; i <= n; ++i)
dp[i][1] = 1;
for(j = 1; j <= m; ++j)
dp[1][j] = 1;
for(i = 2; i <= n; ++i)
for(j = 2; j <= m; ++j)
{
if(mat[i][j-1] != mat[i-1][j] || mat[i][j] == mat[i-1][j] || mat[i][j] == mat[i][j-1])
dp[i][j] = 1;
else
{
int k = dp[i][j] = min(dp[i-1][j],dp[i][j-1]);
if(mat[i][j] == mat[i-k][j-k])
++dp[i][j];
}
ans = max(ans,dp[i][j]);
}
return ans * 4;
}
int main()
{
int t,j,i;
scanf("%d",&t);
int cas = 0;
char ss;

while(t--)
{
scanf("%d%d",&n,&m);
for(i = 1; i <= n; ++i)
for(j = 1; j <= m; ++j)
{
cin >> ss;
mat[i][j] = (ss == 'R');
s[i-1][j-1] = ss;
}
int ans = max(f(1),f(0));
ans = max(ans,dd());
printf("Case #%d: %d\n",++cas,ans);
}
return 0;
}
/*
4 4
RBRB
BRBR
RBRB
BRBR
4 4
RRBR
RBRB
BRBR
RBRB

5 5
BRRRR
RRRBR
BBRBR
BRBRB
BBRBR

10 8
BRBBBBBR
RBRRRRRR
BRBBBBBB
RRRRRRRR
RRRRRRRR
RRBRBRBR
BBRBRBRB
RRBRBRBR
BBRBRBRB
RRBRBRBR

11 8
BRBBBBBR
RBRRRRRR
BRBBBBBB
RRRRRRRR
RRRRRRRR
RRBRBRBR
BBRBRBRB
RRBRBRBR
BBRBRBRB
RRBRBRBR
BBRBRBRB
*/