HDU 1429 Victory Grand Escape (continued)

Speech

Do a lot of search for the basic problem, today began to brush the search of advanced topics, become a search for the great God.

Problem analysis

The meaning of the topic I will not say that you can certainly understand, here the main solution to pick the key and open the door, pay attention here to pick up the key so before we can go back to walk, because the previous road may have not opened the door, then how to say, here to use the state compression.

State compression means that the state is represented by a number, and here I use 2 notation, for example, 1001 means I got the first key and the 4th key, 1101 means I got the first, the third, the fourth key. Therefore, the question of the judgment of the reuse of vis[20][20][1024], the first 2-dimensional representation of position coordinates, the latter one represents the number of keys obtained, the specific look at the code will understand.

#include <cstdio>#include <queue>#include <algorithm>#include <iostream>#include <cstring>using namespace STD;Const intMAXN = -;intN,m,limit;//Maze length, width, and time limitstructnode{intX,y,t,state; Node () {} node (intAintBintCintd): X (a), Y (b), T (c), state (d) {}//Construction};BOOLvis[maxn][maxn][1<<Ten];//3, the last dimension to collect the number of keys, a total of 9 keys, so need to 1<<10;CharMAZE[MAXN][MAXN];intdir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};intBFsintSxintSY) { queue <Node>Que Node Pre;memset(Vis,false,sizeof(VIS)); Que.push (SX, SY,0,0)); vis[sx][sy][0] =1;intState while(!que.empty ()) {pre = Que.front (); Que.pop ();if(Maze[pre.x][pre.y] = =' ^ ')//Search to the end{if(pre.t >= limit)//exceeding limit                return-1;Else                returnPRE.T; } for(inti =0; I <4; i++) {intxx = pre.x + dir[i][0];intyy = Pre.y + dir[i][1];if(XX <0|| XX >= N | | yy <0|| yy >= m | | MAZE[XX][YY] = =' * ')//Judging if it is out of bounds and whether it is a wall                Continue;if(Maze[xx][yy] <=' Z '&& Maze[xx][yy] >=' A ')//Maze This position is the key{state = pre.state| (1<< (Maze[xx][yy]-' A '));//Pick up the key, which is 2 binary and |                if(!vis[xx][yy][state]) {Vis[xx][yy][state] =1; Que.push (Node (xx, yy, pre.t+1, state)); }            }Else if(Maze[xx][yy] <=' Z '&& Maze[xx][yy] >=' A ')//Maze the position for the door, note that as long as you have the key, the door can always be opened{state = pre.state& (1<< (Maze[xx][yy]-' A '));//Whether the key is 2 in the binary and, if not, the value is 0                if(State &&!vis[xx][yy][pre.state]) {Vis[xx][yy][pre.state] =1; Que.push (Node (xx,yy,pre.t+1, pre.state)); }            }Else if(!vis[xx][yy][pre.state])//Maze here for road, start, or end{Vis[xx][yy][pre.state] =1; Que.push (Node (xx, yy, pre.t+1, pre.state)); }        }    }return-1;}intMain () { while(scanf(" %d%d%d", &n, &m, &limit)! = EOF) { for(inti =0; I < n; i++)scanf('%s ', Maze[i]);intSx,sy; for(inti =0; I < n; i++) for(intj =0; J < M; J + +)if(Maze[i][j] = =' @ 'SX = i, sy = j;printf("%d\n", BFS (SX, SY)); }return 0;}

HDU 1429 Victory Grand Escape (continued)