HDU - 1429 - 胜利大逃亡(续)

胜利大逃亡(续)

题目传送：HDU - 1429 - 胜利大逃亡(续)

基础广搜题，注意是用位运算来判断的是否有钥匙的。

少写一个字母找了好久的bug。。好久的bug。。的bug。。bug。。g

AC代码：

#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <cctype>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x7fffffff
using namespace std;

int n, m, t;

bool vis[21][21][1 << 10];

char mp[25][25];

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

struct node {
    int x, y, t, state;
    node() {}
    node(int _x, int _y, int _t, int _state) : x(_x), y(_y), t(_t), state(_state) {}
};

int bfs(int x, int y) {
    memset(vis, false, sizeof(vis));

    queue<node> q;
    q.push(node(x, y, 0, 0));
    vis[x][y][0] = true;

    while(!q.empty()) {
        node cur = q.front();
        q.pop();
//      cout << cur.x << " " << cur.y << " " << cur.state << endl;
        if(cur.t >= t-1) break;
        for(int i = 0; i < 4; i ++) {
            int xx = cur.x + dx[i];
            int yy = cur.y + dy[i];
            int tt = cur.t + 1;
            int sta = cur.state;


            if(xx < 0 || xx >= n || yy < 0 || yy >= m || mp[xx][yy] == ‘*‘ || vis[xx][yy][sta]) {
                continue;
            }

//          cout << cur.x << " " << cur.y << " " << cur.state << endl;

            if(isupper(mp[xx][yy]) && !(sta & (1 << (mp[xx][yy] - ‘A‘))) ) {
                continue;
            }

            if(islower(mp[xx][yy])) {
                sta |= (1 << (mp[xx][yy] - ‘a‘));
            }

            if(mp[xx][yy] == ‘^‘) {
                return tt;
            }

            if(!vis[xx][yy][sta]) {
                vis[xx][yy][sta] = true;
                q.push(node(xx, yy, tt, sta));
            }
        }
    }
    return -1;
}

int main() {
    while(scanf("%d %d %d", &n, &m, &t) != EOF) {
        for(int i = 0; i < n; i ++ ){
            scanf("%s", mp[i]);
        }

        int sx, sy;
        for(int i = 0; i < n; i ++) {
            for(int j = 0; j < m; j ++) {
                if(mp[i][j] == ‘@‘) {
                    sx = i;
                    sy = j;
                    break;
                }
            }
        }

        printf("%d\n", bfs(sx, sy));

    }
    return 0;
}