ZOJ 1649 Rescue(有敌人迷宫BFS)

来源：互联网发布：js map存值编辑：程序博客网时间：2024/05/16 09:06

题意求迷宫中从a的位置到r的位置需要的最少时间经过'.'方格需要1s 经过‘x’方格需要两秒 '#'表示墙

由于有1s和2s两种情况需要在基础迷宫bfs上加些判断

令到达每个点的时间初始为无穷大当从一个点到达该点用的时间比他本来的时间小时更新这个点的时间并将这个点入队扫描完全图就得到答案咯

#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;}

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