题意:给定一个机器人的行走路线,求最少的点能使得机器人可以走这样的路线。
析:每次行走,记录一个方向向量,每次只有是相反方向时,才会增加一个点,最后再加上最后一个点即可。
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 debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<double, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e4 + 10; const int mod = 1e6; 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 bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int main(){ map<char, int> mp; mp['L'] = 0; mp['U'] = 1; mp['D'] = 2; mp['R'] = 3; string s; cin >> n >> s; if(1 == n){ cout << "1 "; return 0; } int ans = 0; int dx = 0, dy = 0, pos = 0; for(int i = 0; i < s.size(); ++i){ if(s[i] == 'U'){ if(dy >= 0) dy = 1; else ++ans, dy = 1, dx = 0; } else if(s[i] == 'D'){ if(dy <= 0) dy = -1; else ++ans, dy = -1, dx = 0; } else if(s[i] == 'L'){ if(dx <= 0) dx = -1; else ++ans, dx = -1, dy = 0; } else{ if(dx >= 0) dx = 1; else ++ans, dx = 1, dy = 0; } } ++ans; printf("%d ", ans); return 0; }