HDU 1429 胜利大逃亡(续) (BFS )

原题链接

题目大意

一个迷宫，相邻房间可能有门，可能有墙，可能直接连通。门的钥匙遍布在各个房间，对应的钥匙打开对应的门。问能否在规定时间之内逃出迷宫？

题目分析

求逃出迷宫最短时间，BFS.

这里存在的问题是：

怎么描述在房间拥有的钥匙的状态？

我们可以用一个整数的各个二进制位来表示是否拥有对应种类钥匙。和拯救大兵瑞恩一样的思路。

AC代码

#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
const int maxn = 20 + 5;
const int keyMax = 10 + 5;

char maze[maxn][maxn];
bool vis[maxn][maxn][1 << keyMax];

int n, m, t;
int sx, sy, ex, ey;

int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};

struct node {
int x, y;
int t;
int curKeys;
};

int bfs() {
queue<node> q;
node s;
s.x = sx, s.y = sy, s.t = 0, s.curKeys = 0;
q.push(s);
while(!q.empty()) {
node cur = q.front(); q.pop();
if (cur.x == ex && cur.y == ey) return cur.t;
for(int i = 0; i < 4; i++) {
int cx = cur.x + dir[i][0];
int cy = cur.y + dir[i][1];
if (cx >= 1 && cx <= n && cy >= 1 && cy <= m) {
if (maze[cx][cy] == '*') continue;
if (maze[cx][cy] >= 'a' && maze[cx][cy] <= 'j') {
node next;
next.x = cx, next.y = cy;
next.t = cur.t + 1;
next.curKeys = cur.curKeys | (1 << (maze[cx][cy] - 'a' + 1));
if (vis[cx][cy][next.curKeys]) vis[cx][cy][next.curKeys] = false, q.push(next);
} else if (maze[cx][cy] >= 'A' && maze[cx][cy] <= 'J'){
if ((cur.curKeys >> (maze[cx][cy] - 'A' + 1)) & 1) {
node next;
next.x = cx, next.y = cy;
next.t = cur.t + 1;
next.curKeys = cur.curKeys;
if (vis[cx][cy][next.curKeys]) vis[cx][cy][next.curKeys] = false, q.push(next);
}
} else {
node next;
next.x = cx, next.y = cy;
next.t = cur.t + 1;
next.curKeys = cur.curKeys;
if (vis[cx][cy][next.curKeys]) vis[cx][cy][next.curKeys] = false, q.push(next);
}
}
}
}
return -1;
}

int main() {
while(~(scanf("%d%d%d", &n, &m, &t))) {
memset(maze, 0, sizeof(maze));
memset(vis, true, sizeof(vis));
getchar();
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
scanf("%c", &maze[i][j]);
if (maze[i][j] == '@') {sx = i; sy = j;}
if (maze[i][j] == '^') {ex = i; ey = j;}
}
getchar();
}
int ans = bfs();
ans < t ? cout << ans : cout << -1;
cout << endl;
}
}