UVaLive 6802 Turtle Graphics (水题，模拟)

题意：给定一个坐标，和一行命令，按照命令走，问你有多少点会被访问超过一次。

析：很简单么，按命令模拟就好，注意有的点可能走了多次，只能记作一次。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std;
 
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e2 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {1, 0, -1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn][maxn];
char s[maxn];
 
int main(){
    int T;  cin >> T;
    int x, y;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d %d", &y, &x);
        scanf("%s", s);
        n = strlen(s);
        int i = 0, j = 0;
        memset(a, 0, sizeof a);
        a[x][y] = 1;
        int ans = 0;
 
        while(i < n){
            if(s[i] == 'F'){
                x += dr[j];
                y += dc[j];
                if(a[x][y] == 1)  ++ans, a[x][y] = 2;
                else if(!a[x][y]) a[x][y] = 1;
            }
            else if(s[i] == 'L')  j = (j+3) % 4;
            else j = (j+1) % 4;
            ++i;
        }
        printf("Case #%d: %d %d %d\n", kase, y, x, ans);
 
    }
    return 0;
}