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 ", kase, y, x, ans); } return 0; }