hdu 1429 bfs+状态压缩

//开一个三维标记数组，标记Ignatius有没有以拿到钥匙的状态到达该位置

//由于钥匙的个数最多十个，所以可以用状态压缩来做

//用1和0表示有没有第i种钥匙，这样对于这个人拿到的钥匙状态就可以用二进制数表示

//用bfs找到最小值

#include<iostream>

#include<cstdio>

#include<cstring>

#include<queue>

using namespace std ;

const int maxn = 30;

const int inf = 0x7fffffff;

int vis[maxn][maxn][1<<10];

char map[maxn][maxn] ;

int ans = ans;

struct node

{

int x , y;

int step ;

int state ;

};

int st_x , st_y ;

int n , m;

int dx[4] = {-1,0,1,0} ;

int dy[4] = {0,1,0,-1} ;

queue<struct node> que;

void bfs()

{

while(que.size())

que.pop() ;

memset(vis, 0 ,sizeof(vis)) ;

struct node first = {st_x , st_y , 0 ,0} ;

vis[st_x][st_y][0] = 1;

que.push(first) ;

while(que.size())

{

struct node now = que.front() ;

que.pop() ;

if(map[now.x][now.y] == '^')

{

ans = now.step ;

break;

}

for(int i = 0;i < 4;i++)

{

struct node next ;

next.x = now.x + dx[i] ;

next.y = now.y + dy[i] ;

if(map[next.x][next.y] == '*' || next.x < 1 || next.x > n ||next.y < 1 || next.y > m)

continue ;

if(map[next.x][next.y] >= 'A' && map[next.x][next.y] <= 'J')

if(!(now.state & (1<<(map[next.x][next.y] - 'A'))))

continue ;

if(map[next.x][next.y] >= 'a' && map[next.x][next.y] <= 'j')

{

next.state = now.state | (1<<(map[next.x][next.y]-'a')) ;

next.step = now.step + 1;

if(vis[next.x][next.y][next.state])

continue ;

que.push(next) ;

vis[next.x][next.y][next.state] = 1;

continue;

}

next.state = now.state ;

next.step = now.step + 1;

if(vis[next.x][next.y][next.state])

continue ;

vis[next.x][next.y][next.state] = 1;

que.push(next) ;

}

}

}

int main()

{

// freopen("in.txt","r",stdin) ;

int time ;

while(scanf("%d%d%d", &n , &m , &time) != EOF)

{

for(int i = 1;i <= n;i++)

{

scanf("%s" , &map[i][1]);

for(int j = 1;j <= m;j++)

if(map[i][j] == '@')

st_x = i,st_y = j;

}

ans = inf ; //可能到不了终点，所以有可能bfs没有更新ans

bfs(); //所以ans每次都要初始化

if(ans < time)

printf("%d\n", ans);

else

printf("-1\n");

}

return 0;

}