昨天花了一整天的时间研究你这个恺撒的问题,写了两个程序,
第一个比第二个要好,试鉴赏.(已调试,未做解释,见谅)
#include "stdio.h"
#include "stdlib.h"
#include "process.h"
#define SUCCESS 1
#define FAIL 0
#define YES 1
#define NO 0


void pseek(int*,int*,int,int,int,int,int);

int sum;

void main(void)
{
int m,n,p,*plane,i,j,k,x,y,t;
int *isseek;
FILE*fin,*fout;
if(!(fin=fopen("f:\\ty.txt","r")))
{
  printf("f:\\ty.txt cann't be found\n");
  exit(0);
}
if(!(fout=fopen("f:\\tx.txt","w")))
{
  printf("f:\\tx.txt cann't be found\n");
  exit(0);
}
rewind(fin);
fscanf(fin,"%d %d %d",&n,&m,&p);
plane=(int*)malloc(m*n*sizeof(int));
isseek=(int*)malloc(m*n*sizeof(int));
for(i=0;i<n;i++)
for(j=0;j<m;j++)
fscanf(fin,"%d",&plane[m*i+j]);
for(i=0;i<p;i++)
{
  x=y=sum=0;
  for(t=0;t<n;t++)
  for(j=0;j<m;j++)
  isseek[m*t+j]=0;
  fscanf(fin,"%d",&k);
  pseek(isseek,plane,n,m,k,x,y);
  fprintf(fout,"%d\n",sum);
}
}

void pseek(int*isseek,int*plane,int n,int m,int k,int x,int y)
{
int result=SUCCESS;
int colx=x,linx=x,liny=y,coly=y;
int i,j;
for(i=x;i<x+k;i++)
for(j=y;j<y+k;j++)
if(!plane[m*i+j])
{
  result=FAIL;
  if(j>coly){colx=i;coly=j;}
  if(i>linx){linx=i;liny=j;}
}

if(result==SUCCESS)sum++;
coly++;
linx++;
isseek[m*x+y]=YES;
if(colx+k<=n&&coly+k<=m&&isseek[m*colx+coly]==NO)
pseek(isseek,plane,n,m,k,colx,coly);
if(linx+k<=n&&liny+k<=m&&isseek[m*linx+liny]==NO)
pseek(isseek,plane,n,m,k,linx,liny);
}

输入文件(f:\ty.txt)的形式:
5 5 2
1 0 1 1 1
1 1 1 1 1
1 1 1 0 1
1 1 1 0 1
1 1 1 1 1
2
3
(f:\tx.txt)输出结果:
8
2
输入文件要注意空格.
#include "stdio.h"
#include "stdlib.h"
#include "process.h"
#define SUCCESS 1
#define FAIL 0
#define YES 1
#define NO 0

void pseek(int*,int*,int,int,int,int,int);

int sum;

void main(void)
{
int m,n,p,*isseek,*plane,i,j,t,k,x,y;
FILE*fin,*fout;
if(!(fin=fopen("f:\\ty.txt","r")))
{
  printf("f:\\ty.txt cann't be found\n");
  exit(0);
}
if(!(fout=fopen("f:\\tx.txt","w")))
{
  printf("f:\\tx.txt cann't be found\n");
  exit(0);
}
rewind(fin);
fscanf(fin,"%d %d %d",&n,&m,&p);
plane=(int*)malloc(m*n*sizeof(int));
isseek=(int*)malloc(m*n*sizeof(int));
for(i=0;i<n;i++)
for(j=0;j<m;j++)
fscanf(fin,"%d",&plane[m*i+j]);
for(i=0;i<p;i++)
{
  sum=x=y=0;
  for(t=0;t<n;t++)
  for(j=0;j<m;j++)
  isseek[m*t+j]=NO;
  fscanf(fin,"%d",&k);
  pseek(isseek,plane,n,m,k,x,y);
  fprintf(fout,"%d\n",sum);
}
}

void pseek(int*isseek,int*plane,int n,int m,int k,int x,int y)
{
int result=SUCCESS;
int colx=x,linx=x,liny=y,coly=y;
int i,j;
for(i=x;i<x+k;i++)
{
for(j=y;j<y+k;j++)
{
if(!plane[m*i+j])
{
  result=FAIL;
break;
}
}
if(result==FAIL)break;
}

if(result==SUCCESS)sum++;
coly++;
linx++;
isseek[m*x+y]=YES;
if(colx+k<=n&&coly+k<=m&&isseek[m*colx+coly]==NO)
pseek(isseek,plane,n,m,k,colx,coly);
if(linx+k<=n&&liny+k<=m&&isseek[m*linx+liny]==NO)
pseek(isseek,plane,n,m,k,linx,liny);
}




来一点注释不行么？本人C水平有限，我学Pascal~~~~~~~~~

很明显,我用了递归算法.
我的算法是:
先搜索一个p*p块,然后则分俩路,一横一竖,继续搜索.
search()
{
若无寺庙
{
横的新块左上角x坐标+1,y坐标不变.search();
竖的新块左上角y坐标+1,x坐标不变.search();
}
若有寺庙
{
找出块内x坐标最大的寺庙,横的新块的x坐标即使改寺庙的x坐标+1,y不变.search();
找出块内y坐标最大的寺庙,竖的新块的y坐标即使改寺庙的y坐标+1,y不变.search();
}
}

当然你很容易发现搜索中可能会出现块的重复搜索.
于是我用了标记数组isseek;

本人解说能力有限,若不懂,吾已无可奈何.