原文链接: HDU 1429 bfs 状态压缩

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题意：迷宫中，一个起点，一个终点，迷宫中有墙，有门，门的钥匙也在迷宫中某处，只有拿到钥匙才能打开门，问能不能再T步（不含）之内逃出迷宫。

题解：在朴素BFS上增加了钥匙的状态，只有有钥匙才能打开门，总共有不超过10吧钥匙，所以用一个int的整数的二进制即可存储钥匙的状态。碰到门先判断状态，碰到钥匙更新状态

字符串后面有\n,最大长度是21

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>#define LL long long
int const MAX = 1e6 + 1;
int const INF = 1 << 30;
double const EPS = 0.00000001;
using namespace std;char g[21][21];
int n, m, T;
bool vis[20][20][1 << 10];
int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};struct Node {int x, y, state, step;
};
Node st, t;void setVis(Node &nd){vis[nd.x][nd.y][nd.state] = 1;
}int bfs(){memset(vis, 0, sizeof(vis));queue<Node> q;q.push(st);setVis(st);while (!q.empty()){Node nd = q.front();q.pop();for (int i = 0; i < 4; i++){t.x = nd.x + dir[i][0], t.y = nd.y + dir[i][1];t.step = nd.step + 1, t.state = nd.state;//只将满足条件的加入队列if (t.x < 0 || t.y < 0 || t.x >= n || t.y >= m ||g[t.x][t.y] == '*' || t.step >= T)continue;char c = g[t.x][t.y];if (c == '^'){return t.step;}//如果是门且没有对应的钥匙if (c >= 'A' && c <= 'J' && !(t.state & (1 << (c - 'A'))))continue;//如果是钥匙,更新状态if (c >= 'a' && c <= 'j')t.state |= (1 << (c - 'a'));if (!vis[t.x][t.y][t.state]){q.push(t);setVis(t);}}}return -1;
}
int main(){freopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);while (scanf("%d%d%d", &n, &m, &T) == 3){for (int i = 0; i < n; i++){scanf("%s", g[i]);for (int j = 0; j < m; j++)if (g[i][j] == '@')st.x = i, st.y = j, st.step = 0, st.state = 0;}printf("%d\n", bfs());}return 0;
}