POJ-2251 Dungeon Master(简单的三维bfs)
2017-07-23 21:42
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Dungeon Master
You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides. Is an escape possible? If yes, how long will it take? Input The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size). L is the number of levels making up the dungeon. R and C are the number of rows and columns making up the plan of each level. Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C. Output Each maze generates one line of output. If it is possible to reach the exit, print a line of the form Escaped in x minute(s). where x is replaced by the shortest time it takes to escape. If it is not possible to escape, print the line Trapped! Sample Input 3 4 5S.....###..##..###.#############.####...###########.#######E1 3 3S###E####0 0 0 Sample Output Escaped in 11 minute(s).Trapped! Source Ulm Local 1997 |
#include <iostream> #include<stdio.h> #include<queue> #include<string.h> using namespace std; const int SIZE=35; int l,r,c; char mp[SIZE][SIZE][SIZE]; int step[SIZE][SIZE][SIZE]; int check[SIZE][SIZE][SIZE]; int to[6][3]={{0,0,1},{0,0,-1},{0,1,0},{0,-1,0},{1,0,0},{-1,0,0}}; struct node { int i,j,k; node(int a,int b,int c) { i=a;j=b;k=c; } }; void bfs(int r1,int r2,int r3,int e1,int e2,int e3) { queue<node> q; while(!q.empty 4000 ()) { q.pop(); } q.push(node(r1,r2,r3)); check[r1][r2][r3]=1; step[r1][r2][r3]=0; while(!q.empty()) { int n1=q.front().i; int n2=q.front().j; int n3=q.front().k; q.pop(); if(n1==e1 && n2==e2 &&n3==e3) break; for(int i=0;i<6;i++) { int new1=n1+to[i][0]; int new2=n2+to[i][1]; int new3=n3+to[i][2]; if(new1>=0 && new1<l &&new2>=0 && new2<r &&new3>=0 && new3<c &&mp[new1][new2][new3]=='.'&&(!check[new1][new2][new3])) { check[new1][new2][new3]=1; step[new1][new2][new3]=step[n1][n2][n3]+1; q.push(node(new1,new2,new3)); } } } } int main() { int r1,r2,r3; int e1,e2,e3; while(cin>>l>>r>>c &&(l||r||c)) { memset(mp,0,sizeof(mp)); memset(step,0,sizeof(step)); memset(check,0,sizeof(check)); for(int i=0;i<l;i++,getchar()) { for(int j=0;j<r;j++,getchar()) { for(int k=0;k<c;k++) { cin>>mp[i][j][k]; if(mp[i][j][k]=='S') { r1=i;r2=j;r3=k; } if(mp[i][j][k]=='E') { e1=i;e2=j;e3=k;mp[i][j][k]='.';//这个地方把最后的E改成.不然最后一步搜不到。 } } } } bfs(r1,r2,r3,e1,e2,e3); if(step[e1][e2][e3]) { printf("Escaped in %d minute(s).\n",step[e1][e2][e3]); } else { puts("Trapped!"); } } return 0; }
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