题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=91
解题思路：BFS搜索
1 /////////////////////////////////////////////////////////////////////////// 2 //problem_id: zoj 1091 3 //user_id: SCNU20102200088 4 /////////////////////////////////////////////////////////////////////////// 5 6 #include <algorithm> 7 #include <iostream> 8 #include <iterator> 9 #include <iomanip> 10 #include <cstring> 11 #include <cstdlib> 12 #include <string> 13 #include <vector> 14 #include <cstdio> 15 #include <cctype> 16 #include <cmath> 17 #include <queue> 18 #include <stack> 19 #include <list> 20 #include <set> 21 #include <map> 22 using namespace std; 23 24 /////////////////////////////////////////////////////////////////////////// 25 typedef long long LL; 26 const double PI=acos(-1.0); 27 28 const int x4[]={-1,0,1,0}; 29 const int y4[]={0,1,0,-1}; 30 const int x8[]={-1,-1,0,1,1,1,0,-1}; 31 const int y8[]={0,1,1,1,0,-1,-1,-1}; 32 33 typedef int T; 34 T max(T a,T b){ return a>b? a:b; } 35 T min(T a,T b){ return a<b? a:b; } 36 /////////////////////////////////////////////////////////////////////////// 37 38 /////////////////////////////////////////////////////////////////////////// 39 //Add Code: 40 int Ex,Ey,Min; 41 bool flag[10][10]; 42 const int xx[]={-2,-1,1,2,2,1,-1,-2}; 43 const int yy[]={1,2,2,1,-1,-2,-2,-1}; 44 45 struct Node{ 46 int x,y,step; 47 Node(int i,int j,int s):x(i),y(j),step(s){} 48 }; 49 50 queue<Node> q; 51 52 void BFS(int i,int j){ 53 if(i==Ex && j==Ey){ 54 Min=0; 55 return ; 56 } 57 memset(flag,0,sizeof(flag)); 58 flag[i][j]=1; 59 Node node(i,j,0); 60 q.push(node); 61 while(!q.empty()){ 62 Node temp=q.front(); 63 q.pop(); 64 for(int p=0;p<8;p++){ 65 int x=temp.x+xx[p],y=temp.y+yy[p]; 66 if(x<1 || x>8 || y<1 || y>8 || flag[x][y]) continue; 67 int step=temp.step+1; 68 flag[x][y]=1; 69 if(x==Ex && y==Ey){ 70 Min=step; 71 while(!q.empty()) q.pop(); 72 return ; 73 } 74 else{ 75 Node res(x,y,step); 76 q.push(res); 77 } 78 } 79 } 80 } 81 /////////////////////////////////////////////////////////////////////////// 82 83 int main(){ 84 /////////////////////////////////////////////////////////////////////// 85 //Add code: 86 char s[5],e[5]; 87 while(scanf("%s%s",s,e)!=EOF){ 88 int Sx=s[0]-'a'+1,Sy=s[1]-'0'; 89 Ex=e[0]-'a'+1,Ey=e[1]-'0'; 90 BFS(Sx,Sy); 91 printf("To get from %s to %s takes %d knight moves. ",s,e,Min); 92 } 93 /////////////////////////////////////////////////////////////////////// 94 return 0; 95 } 96 97 /////////////////////////////////////////////////////////////////////////// 98 /* 99 Testcase: 100 Input: 101 e2 e4 102 a1 b2 103 b2 c3 104 a1 h8 105 a1 h7 106 h8 a1 107 b1 c3 108 f6 f6 109 Output: 110 To get from e2 to e4 takes 2 knight moves. 111 To get from a1 to b2 takes 4 knight moves. 112 To get from b2 to c3 takes 2 knight moves. 113 To get from a1 to h8 takes 6 knight moves. 114 To get from a1 to h7 takes 5 knight moves. 115 To get from h8 to a1 takes 6 knight moves. 116 To get from b1 to c3 takes 1 knight moves. 117 To get from f6 to f6 takes 0 knight moves. 118 */ 119 ///////////////////////////////////////////////////////////////////////////