模拟题。按题目要求编码即可，每过一分钟，更新一下农夫和奶牛的状态。如果该状态以前出现过，说明有循环，不可能到达，输出0.
#include <iostream>
#include <fstream>
using namespace std;
ifstream fin("ttwo.in");
ofstream fout("ttwo.out");
#ifdef _DEBUG
#define out cout
#define in cin
#else
#define out fout
#define in fin
#endif
char grid[10][10];
bool visited[10][10][4][10][10][4];
int north = 0;
int east = 1;
int south = 2;
int west = 3;
//方向为0,1,2,3时向前走一步row和column的增加值
int delr[4] = {
-1,0,1,0
};
int delc[4] = {
0,1,0,-1
};
// farmer的当前row,column,direction.
int fc,fr,fd;
int cc,cr,cd;
bool available(int i,int j)
{
if(i>=0&&i<=9&&j>=0&&j<=9&&grid[i][j]!='*')
return true;
else
return false;
}
void one_step()
{
if( available( fr+delr[fd],fc+delc[fd] ) ){
fr += delr[fd];
fc += delc[fd];
}else{
fd+=1;
fd%=4;
}
if( available( cr+delr[cd],cc+delc[cd] ) ){
cr += delr[cd];
cc += delc[cd];
}else{
cd+=1;
cd%=4;
}
}
void solve()
{
for(int i=0;i<10;++i)
for(int j=0;j<10;++j){
in>>grid[i][j];
if(grid[i][j]=='F')
fr=i,fc=j,fd=0;
if(grid[i][j]=='C')
cr=i,cc=j,cd=0;
}
memset(visited,0,sizeof(visited));
visited[fc][fr][fd][cc][cr][cd] = true;
int res = 0;
while(true){
one_step();
res++;
if( fc==cc&&fr==cr ){
out<<res<<endl;
exit(0);
}
if(visited[fc][fr][fd][cc][cr][cd]){
out<<0<<endl;
exit(0);
}
visited[fc][fr][fd][cc][cr][cd]=true;
}
}
int main(int argc,char *argv[])
{
solve();
return 0;
}