748C Santa Claus and Robot

748C Santa Claus and Robot

// Problem: C. Santa Claus and Robot// Contest: Codeforces - Technocup 2017 - Elimination Round 3// URL: Memory Limit: 256 MB// Time Limit: 2000 ms// 2022-03-14 12:22:38// // Powered by CP Editor (namespace std;#define rep(i,l,r) for(int i=(l);i<=(r);i++)#define per(i,l,r) for(int i=(l);i>=(r);i--)#define ll long long#define mset(s,t) memset(s,t,sizeof(t))#define mcpy(s,t) memcpy(s,t,sizeof(t))#define fi first#define se second#define pb push_back#define all(x) (x).begin(),(x).end()#define SZ(x) ((int)(x).size())#define mp make_pairtypedef pair pii;typedef pair pll;typedef vector vi; typedef vector Vll; typedef vector > vpii;typedef vector > vpll; const ll mod = 1e9 + 7;//const ll mod = 998244353;const double pi = acos(-1.0);inline ll qmi (ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans = ans * a%mod; a = a * a %mod; b >>= 1; } return ans;}inline int read () { int x = 0, f = 0; char ch = getchar(); while (!isdigit(ch)) f |= (ch=='-'),ch= getchar(); while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar(); return f?-x:x;}template void print(T x) { if (x < 0) putchar('-'), x = -x; if (x >= 10) print(x/10); putchar(x % 10 + '0');}inline ll sub (ll a, ll b) { return ((a - b ) %mod + mod) %mod;}inline ll add (ll a, ll b) { return (a + b) %mod;}inline ll inv (ll a) { return qmi(a, mod - 2);}void solve() { int n; cin >> n; map st; st['R'] = st['L'] = 1; st['U'] = st['D'] = 0; string s; cin >> s; int ans = 1; int dx = 0, dy = 0; int mx = 0, my = 0; for (auto t : s) { dx = dy = 0; if (t == 'R') dx = 1; if (t == 'L') dx = -1; if (t == 'U') dy = -1; if (t == 'D') dy = 1; if (dx * mx == -1 || dy * my == -1) { ans ++; mx = my = 0; } if (mx == 0) mx = dx; if (my == 0) my = dy; } cout << ans << endl;}int main () { // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0); int t; t =1; //cin >> t; while (t --) solve(); return 0;}

题意:给你一个操作序列，问最少有多少个点？ 从贪心的角度想。题目中说每次走最短路。因此如果存在两个相反方向的走法，那么肯定不是最短路，因此需要产生一个点

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