【HDOJ】1253 胜利大逃亡

经典的BFS，需要注意的是当前时间超过最小时间，输出-1。同时，队列为空时还未返回，证明并未找到终点（可能终点为墙）。此时也应该输出-1，这个部分容易wa。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
using namespace std;

#define MAXNUM 50
#define SETPOS(pos,xx,yy,zz,tt) pos.x=xx;pos.y=yy;pos.z=zz;pos.t=tt;

typedef struct {
    int x, y, z;
    int t;
} pos_st;

int map[MAXNUM][MAXNUM][MAXNUM];
int visit[MAXNUM][MAXNUM][MAXNUM];
int direct[6][3] = {{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};
int a, b, c, mintime;
queue<pos_st> poss;

void bfs(int x, int y, int z, int t) {
    pos_st pos, tmp;

    SETPOS(pos,x,y,z,t);
    poss.push(pos);
    visit[x][y][z] = 1;

    while (!poss.empty()) {
        pos = poss.front();
        poss.pop();

        if (pos.t > mintime) {
            printf("-1
");
            return;
        }
        if (pos.x==a-1 && pos.y==b-1 && pos.z==c-1) {
            printf("%d
", pos.t);
            return;
        }
        for (int i=0; i<6; ++i) {
            SETPOS(tmp, pos.x+direct[i][0], pos.y+direct[i][1], pos.z+direct[i][2], pos.t+1);
            if (tmp.x<0 || tmp.x>=a || tmp.y<0 || tmp.y>=b || tmp.z<0 || tmp.z>=c)
                continue;
            if (visit[tmp.x][tmp.y][tmp.z] || map[tmp.x][tmp.y][tmp.z]==1)
                continue;
            visit[tmp.x][tmp.y][tmp.z] = 1;
            poss.push(tmp);
        }
    }

    printf("-1
");
}

int main() {
    int case_n;
    int i, j, k;

    scanf("%d", &case_n);

    while (case_n--) {
        scanf("%d%d%d%d", &a,&b,&c,&mintime);
        for (i=0; i<a; ++i)
            for (j=0; j<b; ++j)
                for (k=0; k<c; ++k)
                    scanf("%d", &map[i][j][k]);
        memset(visit, 0, sizeof(visit));
        while (!poss.empty())
            poss.pop();
        bfs(0,0,0,0);
    }

    return 0;
}