ZOJ 1649 Rescue(有敵人迷宮BFS)，zojbfs

ZOJ 1649 Rescue(有敵人迷宮BFS)，zojbfs

題意 求迷宮中從a的位置到r的位置需要的最少時間  經過'.'方格需要1s  經過‘x’方格需要兩秒  '#'表示牆

由於有1s和2s兩種情況  需要在基礎迷宮bfs上加些判斷

令到達每個點的時間初始為無窮大  當從一個點到達該點用的時間比他本來的時間小時  更新這個點的時間並將這個點入隊  掃描完全圖就得到答案咯

#include<cstdio>#include<cstring>#include<queue>using namespace std;const int N = 205;char mat[N][N];int time[N][N], sx, sy;int dx[4] = {0, 0, -1, 1};int dy[4] = { -1, 1, 0, 0};struct grid{    int x, y;    grid(int xx = 0, int yy = 0): x(xx), y(yy) {}};void bfs(){    memset(time, 0x3f, sizeof(time));    time[sx][sy] = 0;    queue<grid> g;    g.push(grid(sx, sy));    while(!g.empty())    {        grid cur = g.front();        g.pop();        int cx = cur.x, cy = cur.y, ct = time[cx][cy];        for(int i = 0; i < 4; ++i)        {            int nx = cx + dx[i], ny = cy + dy[i];            if(mat[nx][ny] && mat[nx][ny] != '#')            {                int tt = ct + 1;                if(mat[cx][cy] == 'x') ++tt;                if(tt < time[nx][ny])                {                    time[nx][ny] = tt;                    g.push(grid(nx, ny));                }            }        }    }}int main(){    int n, m, ex, ey;    while (~scanf("%d%d", &n, &m))    {        memset(mat, 0, sizeof(mat));        for(int i = 1; i <= n; ++i)            scanf("%s", mat[i] + 1);        for(int i = 1; i <= n; ++i)            for(int j = 1; j <= m; ++j)                if(mat[i][j] == 'a') sx = i, sy = j;                else if(mat[i][j] == 'r') ex = i, ey = j;        bfs();        if(time[ex][ey] != time[0][0])            printf("%d\n", time[ex][ey]);        else            printf("Poor ANGEL has to stay in the prison all his life.\n");    }    return 0;}