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  • CodeForces 748C Santa Claus and Robot (思维)

    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;
    }
    
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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/6505563.html
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