http://acm.hdu.edu.cn/showproblem.php?pid=1429
一个广搜的简单题吧，不过有意思的事这个题目用到了位运算，还有就是很恶心的MLE
1 #include <stdio.h> 2 #include <string.h> 3 #include <queue> 4 using namespace std; 5 6 int m,n,t; 7 char graph[25][25]; 8 bool vis[24][24][1<<10]; 9 int dic[4][2] = {-1,0,1,0,0,-1,0,1}; 10 struct note{ 11 12 int x,y,state,time; 13 14 }loc; 15 queue<note>s; 16 /* 17 位运算就是在下面，对于每一个钥匙用二进制的0,1来存 18 */ 19 int bfs(int x,int y) 20 { 21 while(!s.empty()) 22 s.pop(); 23 loc.state = 0; 24 loc.time = 1; 25 loc.x = x; 26 loc.y = y; 27 s.push(loc); 28 while(!s.empty()) 29 { 30 note fa,son; 31 fa = s.front(); 32 s.pop(); 33 // if(fa.time == t) return -1; 34 for(int i = 0 ; i < 4 ; i++) 35 { 36 son.x = fa.x + dic[ i ][ 0 ]; 37 son.y = fa.y + dic[ i ][ 1 ]; 38 if(son.x < 1 || son.y < 1|| son.y > n || son.x > m||!vis[son.x][son.y][fa.state]) continue; //这一行很关键，没有这一行就是MLE 39 if(graph[son.x][son.y] == '^') 40 { 41 if(fa.time == t ) return -1; 42 else return fa.time; 43 } 44 if(graph[son.x][son.y] >='a'&& graph[son.x][son.y] <='z' &&vis[son.x][son.y][fa.state| (1 <<(graph[son.x][son.y]-'a'))]) 45 { 46 son.state = fa.state | (1 <<(graph[son.x][son.y]-'a')); 47 son.time = fa.time+1; 48 vis[son.x][son.y][son.state] = false; 49 s.push(son); 50 } 51 else if(graph[son.x][son.y]<='Z' && graph[son.x][son.y] >='A' && fa.state & (1 << graph[son.x][son.y]-'A') &&vis[son.x][son.y][fa.state]) 52 { 53 son.state = fa.state; 54 son.time = fa.time+1; 55 vis[son.x][son.y][son.state] = false; 56 s.push(son); 57 } 58 else if(graph[son.x][son.y]=='.'||graph[son.x][son.y]=='@'&&vis[son.x][son.y][fa.state]) 59 { 60 son.state = fa.state; 61 son.time = fa.time + 1 ; 62 vis[son.x][son.y][son.state] = false; 63 s.push(son); 64 } 65 } 66 } 67 return -1; 68 } 69 70 int main() 71 { 72 // freopen("in.txt","r",stdin); 73 int x,y; 74 while(~scanf("%d%d%d",&m,&n,&t)) 75 { 76 getchar(); //吃掉一个换行符 77 memset(graph,0,sizeof(graph)); 78 memset(vis,true,sizeof(vis)); 79 int ans; 80 for(int i = 1 ; i <= m; i++) 81 { 82 for(int j = 1 ; j <= n ; j++) 83 { 84 scanf("%c",&graph[i][j]); 85 if(graph[i][j]=='@') 86 x = i,y = j; 87 } 88 getchar(); 89 } 90 ans = bfs(x,y); 91 printf("%d ",ans); 92 } 93 return 0; 94 }