HDU 1885 Key Task (BFS + 状态压缩)

题意：给定一个n*m的矩阵，里面有门，有钥匙，有出口，问你逃出去的最短路径是多少。

析：这很明显是一个BFS，但是，里面又有其他的东西，所以我们考虑状态压缩，定义三维BFS，最后一维表示拿到钥匙的状态，然后再BFS，就简单了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 5;
const int mod = 1e9 + 7;
const int dr[] = {0, 1, 0, -1};
const int dc[] = {1, 0, -1, 0};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
struct Node{
    int x, y, d, state;
    Node() { }
    Node(int xx, int yy, int dd, int k) : x(xx), y(yy), d(dd), state(k) { }
};
char s[maxn][maxn];
bool vis[maxn][maxn][20];
map<char, int> key, door;

void bfs(int x, int y){
    queue<Node> q;
    memset(vis, 0, sizeof vis);
    q.push(Node(x, y, 0, 0));
    vis[x][y][0] = true;

    while(!q.empty()){
        Node u = q.front(); q.pop();
        for(int i = 0; i < 4; ++i){
            int x = u.x + dr[i];
            int y = u.y + dc[i];
            int d = u.d + 1;
            if(!is_in(x, y))  continue;
            if(s[x][y] == 'X'){
                printf("Escape possible in %d steps.
", d);
                return ;
            }
            if(s[x][y] == '#')  continue;
            int state;
            if(islower(s[x][y])) state = u.state | (1 << key[s[x][y]]);
            else if(isupper(s[x][y])){
                if(u.state & (1<<door[s[x][y]]))  state = u.state;
                else continue;
            }
            else if(s[x][y] == '.' || s[x][y] == '*')  state = u.state;
            if(!vis[x][y][state]){
                vis[x][y][state] = true;
                q.push(Node(x, y, d, state));
            }
        }
    }
    printf("The poor student is trapped!
");
}

int main(){
    key['b'] = 1;  door['B'] = 1;
    key['y'] = 2;  door['Y'] = 2;
    key['r'] = 3;  door['R'] = 3;
    key['g'] = 0;  door['G'] = 0;
    while(scanf("%d %d", &n, &m) == 2){
        if(!n && !m)  break;
        for(int i = 0; i < n; ++i)  scanf("%s", s+i);
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < m; ++j)
                if(s[i][j] == '*') { bfs(i, j);  break;  }
    }
    return 0;
}