USACO Section 5.2 Wisconsin Squares - 按要求DFS就行了..

       这题真搞~就一组数据....囧~~~

       按他的要求枚举搜索就ok了..当然要输出字典序最小的解~~~那就按字典序来搜索..除了最开始是确定D..后面的都是先确定较小的A,B,C,D,E..再确定较小的Y..再确定较小的X...如此搜出的第一个解就是字典序最小的答案...

       开始还想着Hash...结果写完了样例过了...一交就A了...似乎这题不需要Hash..

Program:

/*  ID: zzyzzy12   LANG: C++   TASK: wissqu*/      #include<iostream>      #include<istream>  #include<stdio.h>     #include<string.h>      #include<math.h>      #include<stack>#include<map>#include<algorithm>      #include<queue>   #define oo 2000000005  #define ll long long  #define pi (atan(2)+atan(0.5))*2 using namespace std;char s[6][6];bool used[6][6];int num,have[5];struct node{      char t;      int y,x;     }way[20];bool ok(int y,int x,char c){      if (used[y][x] || s[y][x]==c) return false;      if (s[y][x+1]==c || s[y][x-1]==c || s[y+1][x]==c || s[y-1][x]==c) return false;      if (s[y+1][x+1]==c || s[y+1][x-1]==c ) return false;      if (s[y-1][x+1]==c || s[y-1][x-1]==c) return false;      return true;       } void DFS(int step){      int x,y,i;      char c;      if (step==17)      {             if (!num)             {                    for (i=1;i<=16;i++)                        printf("%c %d %d\n",way[i].t,way[i].y,way[i].x);               }                   num++;             return;             }      for (i=0;i<5;i++)        if (have[i])        {               have[i]--;               for (y=1;y<=4;y++)                 for (x=1;x<=4;x++)                 if (ok(y,x,'A'+i))                 {                        used[y][x]=true;  c=s[y][x];                        s[y][x]='A'+i;                          way[step].t=i+'A'; way[step].y=y; way[step].x=x;                        DFS(step+1);                              s[y][x]=c;        used[y][x]=false;                   }               have[i]++;        }           return;       }int main()  {        freopen("wissqu.in","r",stdin);         freopen("wissqu.out","w",stdout);      int i,j;      char c;      for (i=0;i<=5;i++) s[0][i]=s[i][0]=s[5][i]=s[i][5]='?';      for (i=1;i<=4;i++) scanf("%s",s[i]+1);      num=0;      memset(used,false,sizeof(used));      have[0]=have[1]=have[2]=have[4]=3; have[3]=4;      for (i=1;i<=4;i++)         for (j=1;j<=4;j++)            if (ok(i,j,'D'))            {                  c=s[i][j];   s[i][j]='D';                  used[i][j]=true;  have[3]--;                  way[1].t='D'; way[1].y=i; way[1].x=j;                  DFS(2);  have[3]++;                    s[i][j]=c;    used[i][j]=false;             }      printf("%d\n",num);      return 0;     }