模擬題。按題目要求編碼即可，每過一分鐘，更新一下農夫和奶牛的狀態。如果該狀態以前出現過，說明有循環，不可能到達，輸出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;
//方向為0,1,2,3時向前走一步row和column的增加值
int?delr[4]?=?{
????-1,0,1,0
};
int?delc[4]?=?{
????0,1,0,-1
};
//?farmer的當前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;
}