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Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13741 Accepted Submission(s): 3415
1 /************************************************************************* 2 > File Name: code/hdu/2102.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年10月02日 星期五 09时43分07秒 6 ************************************************************************/ 7 8 #include<iostream> 9 #include<iomanip> 10 #include<cstdio> 11 #include<algorithm> 12 #include<cmath> 13 #include<cstring> 14 #include<string> 15 #include<map> 16 #include<set> 17 #include<queue> 18 #include<vector> 19 #include<stack> 20 #include<cctype> 21 #define y1 hust111qqz 22 #define yn hez111qqz 23 #define j1 cute111qqz 24 #define ms(a,x) memset(a,x,sizeof(a)) 25 #define lr dying111qqz 26 using namespace std; 27 #define For(i, n) for (int i=0;i<int(n);++i) 28 typedef long long LL; 29 typedef double DB; 30 const int inf = 0x3f3f3f3f; 31 char maze[2][11][11]; 32 int d[2][11][11]; 33 int dx[4] = {1,0,-1,0}; 34 int dy[4] = {0,-1,0,1}; 35 bool vis[2][11][11]; 36 int n,m,t; 37 struct node 38 { 39 int x,y,z; 40 int d; 41 bool ok() 42 { 43 if (x<0||y<0||x>=n||y>=m) return false; 44 if (vis[z][x][y]) return false; 45 if (maze[z][x][y]==‘*‘) return false; 46 if (maze[z][x][y]==‘#‘&&(maze[z^1][x][y]==‘#‘||maze[z^1][x][y]==‘*‘||vis[z^1][x][y])) return false; 47 return true; 48 } 49 50 }; 51 node s; 52 bool bfs() 53 { 54 s.x = 0 ; 55 s.y = 0; 56 s.z = 0; 57 s.d = 0 ; 58 queue<node>q; 59 q.push(s); 60 vis[0][0][0] = true; 61 while (!q.empty()) 62 { 63 node pre; 64 pre = q.front(); 65 q.pop(); 66 char tmp = maze[pre.z][pre.x][pre.y]; 67 // cout<<pre.z<<" "<<pre.x<<" "<<pre.y<<endl; 68 if (tmp==‘P‘) 69 { 70 // cout<<"pre.d"<<pre.d<<endl; 71 if (pre.d<=t) return true; 72 return false; 73 } 74 75 76 for ( int i = 0 ; i < 4 ; i++) 77 { 78 node nxt; 79 nxt.x = pre.x + dx[i]; 80 nxt.y = pre.y + dy[i]; 81 nxt.z = pre.z; 82 nxt.d = pre.d + 1; 83 if (!nxt.ok()) continue; 84 if (maze[nxt.z][nxt.x][nxt.y]==‘.‘||maze[nxt.z][nxt.x][nxt.y]==‘P‘) 85 { 86 q.push(nxt); 87 vis[nxt.z][nxt.x][nxt.y]= true; 88 } 89 else 90 { 91 nxt.z = pre.z ^ 1; 92 q.push(nxt); 93 vis[nxt.z][nxt.x][nxt.y] = true; 94 } 95 96 } 97 } 98 // cout<<"ahhhh"<<endl; 99 return false; 100 101 102 } 103 int main() //遇到传送门就一定要传送!并不是可以选择的! 104 { //所以如果一个传送门传送到的位置是障碍物,那这个转送门直接就是不可以到达的!等效于障碍物! 105 #ifndef ONLINE_JUDGE 106 freopen("in.txt","r",stdin); 107 #endif 108 int T; 109 scanf("%d",&T); 110 while (T--) 111 { 112 ms(vis,false); 113 scanf("%d %d %d",&n,&m,&t); 114 for ( int i = 0 ; i < n ; i++) scanf("%s",maze[0][i]); 115 for ( int i = 0 ; i < n ; i++) scanf("%s",maze[1][i]); 116 if (bfs()) 117 { 118 puts("YES"); 119 } 120 else 121 { 122 puts("NO"); 123 } 124 } 125 126 #ifndef ONLINE_JUDGE 127 fclose(stdin); 128 #endif 129 return 0; 130 }
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原文地址:http://www.cnblogs.com/111qqz/p/4852198.html