HDU 1035 Robot Motion 模拟

题意：给你一个地图，机器人按照地图上的指令走，有两种情况，一种：走出地图了，二种：走了一段时间之后一直在走一个圈。就是输出这两种情况。

思路：按照题意模拟。

http://acm.hdu.edu.cn/showproblem.php?pid=1035

/*********************************************
    Problem : HDU 1035
    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-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5+5;
const int MAXE = 2e5+5;

typedef long long LL;
typedef unsigned long long ULL;

int T,n,m,k;
char Map[15][15];
int vis[15][15];
int st , idx , op ;
int fx[] = {0,1,-1,0,0};
int fy[] = {0,0,0,1,-1};
int ok ;

void dfs(int x,int y) {
    if(ok) return;
    vis[x][y] = ++ idx;
    switch(Map[x][y]) {
        case 'W' : op = 4 ; break ;
        case 'E' : op = 3 ; break ;
        case 'N' : op = 2 ; break ;
        case 'S' : op = 1 ; break ;
    }
    int tmpx = x + fx[op] ; 
    int tmpy = y + fy[op] ;
    if(tmpx >= 1 && tmpx <= n && tmpy >= 1 && tmpy <= m) {
        if(vis[tmpx][tmpy]) {
            printf("%d step(s) before a loop of %d step(s)\n",vis[tmpx][tmpy]-1,idx-vis[tmpx][tmpy]+1);
            ok = 1;
            return ;
        }
        else dfs(tmpx,tmpy);
    }
    else {
        printf("%d step(s) to exit\n",idx);
        ok = 1;
        return ;
    }
}

void input() {
    scanf("%d",&st);
    rep(i,1,n) scanf("%s",Map[i]+1) ;//, printf("%s",Map[i]+1);
}

void solve() {
    idx = 0; ok = 0;
    cls(vis,0);
    dfs(1,st);
}

int main(void) {
    //freopen("a.in","r",stdin);
    //scanf("%d",&T); while(T--) {
    while(scanf("%d %d",&n,&m),n+m) {
    //while(scanf("%d",&n),n) {
        input();
        solve();
    }
    return 0;
}