Poj 3083 Children of the Candy Corn

题目连接：http://poj.org/problem?id=1077

题目描述：给出一个迷宫，#表示墙，S表示起点，E表示终点。求从S到E的优先靠左边的墙走、优先靠右边的墙走和S到E的最短路径。

解法：求优先靠左靠右的路径，用dfs，求最短路径用bfs即可。

dfs中可以用nx = x+move[(i+4)%4][0];ny = y+move[(i+4)%4][1];处理很巧妙

View Code

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cmath>
  4 #include<cstdio>
  5 #include<queue>
  6 #define see(x) cout<<#x<<":"<<x<<endl;
  7 using namespace std;
  8 const int inf = 9999999;
  9 const int N = 50;
 10 char map[N][N];
 11 typedef struct{
 12     int x, y, dis;
 13 }Node;
 14 
 15 const int move[4][4] ={{0,-1},{-1,0},{0,1},{1,0}};
 16 bool vis[N][N] = {0};
 17 int r, c;
 18 queue<Node> q;
 19 int bfs(){
 20     int i;
 21     Node temp, temp1;
 22     int minn = inf;
 23     while(!q.empty()){
 24         temp = q.front();
 25         q.pop();
 26         if(map[temp.x][temp.y]=='E'){
 27             if(temp.dis<minn){
 28                 minn = temp.dis;
 29             }
 30         }
 31         
 32         else{
 33             for(i=0;i<4;i++){
 34                 temp1.x = temp.x+move[i][0];
 35                 temp1.y = temp.y+move[i][1];
 36                 temp1.dis = temp.dis+1;
 37                 if(temp1.x>=0&&temp1.x<r&&temp1.y>=0&&temp1.y<c&&map[temp1.x][temp1.y]!='#'&&!vis[temp1.x][temp1.y]){
 38                     vis[temp1.x][temp1.y] = 1;
 39                     q.push(temp1);
 40                 }
 41             }
 42         }
 43     }
 44     return minn;
 45 }
 46 int dfs(int x, int y, int dir, char mark){
 47     int i, ndir, nx, ny;
 48     if(map[x][y]=='E'){
 49         return 1;
 50     }
 51     if(mark=='l'){
 52         for(i=dir-1;i<dir+3;i++){
 53             nx = x+move[(i+4)%4][0];
 54             ny = y+move[(i+4)%4][1];
 55             ndir = (i+4)%4;
 56             if(nx>=0&&nx<r&&ny>=0&&ny<c&&map[nx][ny]!='#'){
 57             //    cout<<"nx: "<<nx<<"   ny: "<<ny<<"    "; see(ndir)
 58                 return dfs(nx,ny,ndir,mark)+1;
 59             }
 60         }
 61     }
 62     if(mark=='r'){
 63         for(i=dir+1;i>dir-3;i--){
 64             nx = x+move[(i+4)%4][0];
 65             ny = y+move[(i+4)%4][1];
 66             ndir = (i+4)%4;
 67             if(nx>=0&&nx<r&&ny>=0&&ny<c&&map[nx][ny]!='#'){
 68                 return dfs(nx,ny,ndir,mark)+1;
 69             }
 70         }
 71     }
 72     return 0;
 73 }
 74 int main(){
 75     int t, lp, rp, sp;
 76     int i, j, k, dir;
 77     Node s;
 78     scanf("%d",&t);
 79     while(t--){
 80         scanf("%d%d",&c,&r);
 81         for(i=0;i<r;i++)
 82             scanf("%s",map[i]);
 83         for(i=0;i<r;i++){
 84             for(j=0;j<c;j++){
 85                 if(map[i][j]=='S'){
 86                     s.x=i; s.y=j; s.dis=1;
 87                     break;
 88                 }
 89             }
 90         }
 91         if(s.x==0){
 92             dir=3;
 93         }
 94         else if(s.x==1){
 95             dir=1;
 96         }
 97         else if(s.y==0){
 98             dir = 2;
 99         }
100         else{
101             dir = 0;
102         }
103         lp = dfs(s.x,s.y,dir,'l');
104         
105         vis[s.x][s.y] = 1;
106         rp = dfs(s.x,s.y,dir,'r');
107         
108         memset(vis,0,sizeof(vis));
109         vis[s.x][s.y] = 1;  s.dis = 1;
110         q.push(s);
111         sp = bfs();
112         printf("%d %d %d\n",lp,rp,sp);
113     }
114     return 0;
115 }