USACO 2.4 The Tamworth Two<简单图问题>

meaning：题意是给一张图，图中有牛的位置，农场主的位子和障碍物。牛和农场主只能沿着一个方向走直到要出界或者走到障碍物就转弯！！（注意是走到障碍物才转弯，而不是碰到之前转弯），转弯的方向为顺时针，问农场主是否可以和牛相遇。

the reason of failure：1、需要注意是走到障碍物才转弯。

2、给的图有C和F，如果只遇到'.'才走下去显然是错的，遇到C||F都可以接着走下去。

3、判定不能走的条件不仅是他们在之前走过的位子，还要在这个位子面对的方向与之前相同。

thinking：直接按着题意模拟走就行，认真看题再码，别着急

代码:

/*ID: mePROG: ttwoLANG: C++*/#include <iostream>#include <stdio.h>#include <string.h>#include <queue>using namespace std;struct ttt{int r[2];int c[2];int d[2];int step;int l[2];};char map1[12][12];int walked[11][11][11][11][11][11];int dd(ttt a,int b){    if(a.r[b]>=1&&a.r[b]<=10&&a.c[b]>=1&&a.c[b]<=10)return 1;    else return 0;}int dd2(char c){    if(c=='.'||c=='F'||c=='C')return 1;    else return 0;}int main(){    freopen("ttwo.in","r",stdin);    freopen("ttwo.out","w",stdout);    int i,j,k,l,f1,f2,f3,t71,t2,t3;    ttt s1,s2;    memset(map1,0,sizeof(map1));    memset(walked,0,sizeof(walked));    for(i=1;i<=10;i++)    for(j=1;j<=10;j++){        cin >> map1[i][j];        if(map1[i][j]=='F'){            s1.r[0]=i;            s1.c[0]=j;        }else if(map1[i][j]=='C'){            s1.r[1]=i;            s1.c[1]=j;        }    }    s1.step=0;    s1.d[0]=0;    s1.d[1]=0;    queue<ttt>qq;    qq.push(s1);    while(!qq.empty()){        s1=qq.front();qq.pop();        //cout << s1.r[0] << "," << s1.c[0] << "-" <<s1.d[0] << "   "        //<<s1.r[1] << "," << s1.c[1] << "-" << s1.d[1];        if(s1.r[0]==s1.r[1]&&s1.c[0]==s1.c[1]){            break;        }        //cout << "   没出去 ";        //if(s1.l[0]==0){        if(s1.d[0]==0){            if(dd2(map1[s1.r[0]][s1.c[0]])&&dd2(map1[s1.r[0]-1][s1.c[0]])){                s1.r[0]--;            }else{                s1.d[0]=(s1.d[0]+1)%4;            }        }else if(s1.d[0]==3){            if(dd2(map1[s1.r[0]][s1.c[0]])&&dd2(map1[s1.r[0]][s1.c[0]-1])){                s1.c[0]--;            }else{                s1.d[0]=(s1.d[0]+1)%4;            }        }else if(s1.d[0]==2){        if(dd2(map1[s1.r[0]][s1.c[0]])&&dd2(map1[s1.r[0]+1][s1.c[0]])){                s1.r[0]++;            }else{                s1.d[0]=(s1.d[0]+1)%4;            }            }else if(s1.d[0]==1){            if(dd2(map1[s1.r[0]][s1.c[0]])&&dd2(map1[s1.r[0]][s1.c[0]+1])){                s1.c[0]++;            }else{                s1.d[0]=(s1.d[0]+1)%4;            }            }            //if(s1.l[1]==0){            if(s1.d[1]==0){            if(dd2(map1[s1.r[1]][s1.c[1]])&&dd2(map1[s1.r[1]-1][s1.c[1]])){                s1.r[1]--;            }else{                s1.d[1]=(s1.d[1]+1)%4;            }        }else if(s1.d[1]==3){            if(dd2(map1[s1.r[1]][s1.c[1]])&&dd2(map1[s1.r[1]][s1.c[1]-1])){                s1.c[1]--;            }else{                s1.d[1]=(s1.d[1]+1)%4;            }        }else if(s1.d[1]==2){        if(dd2(map1[s1.r[1]][s1.c[1]])&&dd2(map1[s1.r[1]+1][s1.c[1]])){                s1.r[1]++;            }else{                s1.d[1]=(s1.d[1]+1)%4;            }            }else if(s1.d[1]==1){            if(dd2(map1[s1.r[1]][s1.c[1]])&&dd2(map1[s1.r[1]][s1.c[1]+1])){                s1.c[1]++;            }else{                s1.d[1]=(s1.d[1]+1)%4;            }            }        //    cout << s1.r[0] << "," << s1.c[0] << "-" <<s1.d[0] << "   "      //  <<s1.r[1] << "," << s1.c[1] << "-" << s1.d[1] << "  " ;    //    cout << walked[s1.r[0]][s1.c[0]][s1.d[0]][s1.r[1]][s1.c[1]][s1.d[1]] << endl;            if(walked[s1.r[0]][s1.c[0]][s1.d[0]][s1.r[1]][s1.c[1]][s1.d[1]]==0){                walked[s1.r[0]][s1.c[0]][s1.d[0]][s1.r[1]][s1.c[1]][s1.d[1]]=s1.step;                s1.step++;  //              cout << " 接着走" <<s1.step << endl;                qq.push(s1);            }    }    //cout << endl;    if(s1.r[0]==s1.r[1]&&s1.c[0]==s1.c[1]){            cout << s1.step << endl;        }else{        cout << 0 <<endl;        }    return 0;}