HDU 1072 Nightmare BFS

题意：在N∗M的矩阵内只可以走六步，如果碰到4就可以再走六步，问能不能走出迷宫？（如果能走出，输出最少的秒数，如果不能走出，输出-1。）

思路：BFS呀，我开始蠢，写了DFS。建一个t数组，tij代表走到(i,j)还剩几步可以走，如果碰到4的话，设置一个时间，重新加入队列即可。

http://acm.hdu.edu.cn/showproblem.php?pid=1072

/*********************************************    Problem : HDU 1072    Author  : NMfloat    InkTime (c) NM . All Rights Reserved .********************************************/#include <map>#include <set>#include <queue>#include <cmath>#include <ctime>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++)#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --)#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)#define cls(a,x)   memset(a,x,sizeof(a))#define eps 1e-8using namespace std;const int MOD = 1e9+7;const int INF = 0x3f3f3f3f;const int MAXN = 1e5+5;const int MAXE = 2e5+5;struct Node {    int x,y,t;//t代表走到这一秒最少的时间    int cost;//消耗的时间    Node(){}    Node(int _x,int _y,int _t,int _cost){x = _x ; y = _y ; t = _t;cost = _cost;}};typedef long long LL;typedef unsigned long long ULL;int T,n,m,k;queue<Node>q;int fx[] = {0,1,-1,0,0};int fy[] = {0,0,0,-1,1};int sx,sy;int Map[10][10];int t[10][10];//走到这一步最短的时间int ok ;void BFS() {    while(!q.empty()) q.pop();    q.push(Node(sx,sy,6,0));    t[sx][sy] = 6;    while(!q.empty()) {        Node tmp = q.front();        q.pop();        // printf("x : %d y : %d tmp : %d cost : %d\n",tmp.x,tmp.y,tmp.t,tmp.cost);        if(tmp.t == 0) continue;        if(Map[tmp.x][tmp.y] == 3) {            ok = 1;            printf("%d\n",tmp.cost);            break;        }        rep(i,1,4) {            int tmpx = tmp.x + fx[i];            int tmpy = tmp.y + fy[i];            int tmpt = tmp.t - 1;            if(tmpx >= 1 && tmpx <= n && tmpy >= 1 && tmpy <= m) {                if(t[tmpx][tmpy] < tmpt) {                    t[tmpx][tmpy] = tmpt;                    if(Map[tmpx][tmpy] == 1 || Map[tmpx][tmpy] == 3) {                        q.push(Node(tmpx,tmpy,tmp.t-1,tmp.cost+1));                    }                    else if(Map[tmpx][tmpy] == 4) {                        t[tmpx][tmpy] = 6;                        q.push(Node(tmpx,tmpy,6,tmp.cost+1));                    }                }            }        }    }}void input() {    scanf("%d %d",&n,&m);    rep(i,1,n) rep(j,1,m) {        scanf("%d",&Map[i][j]);        if(Map[i][j] == 2) {            sx = i ; sy = j;        }    }    }void solve() {    cls(t,0);    ok = 0;    BFS();    if(!ok) puts("-1");}int main(void) {    //freopen("a.in","r",stdin);    scanf("%d",&T); while(T--) {    //while(~scanf("%d %d",&n,&m)) {    //while(scanf("%d",&n),n) {        input();        solve();    }    return 0;}