题意:给定上正方形,圆,三角形,让你求出包围它的最短的路径。
析:首先,如果是这种情况 三角形 三角形 三角形 正方形(圆) 三角形 三角形 三角形 。。这一种就是直接从左边直接连到正方形(圆),也就是相切,剩下的情况都是直接是直线,只要处理一下边界就好。
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> #include <list> #include <assert.h> #include <bitset> #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 fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) //#define sz size() #define pu push_up #define pd push_down #define cl clear() #define all 1,n,1 #define FOR(x,n) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef double lb; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e15; const double inf = 1e20; const lb PI = acos(-1.0); const double eps = 1e-8; const int maxn = 50; const int mod = 7; 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; } char str[maxn]; double Pow(double x){ return x * x; } int main(){ while(scanf("%d", &n) == 1){ scanf("%s", str); int len = strlen(str); int change_pos = len + 1; int pre_T = 0, last_T = 0; for(int i = 0; i < n; i++){ if(str[i] == 'T') pre_T++; else break; } for(int i = n - 1; i >= 0; i--){ if(str[i] == 'T') last_T++; else break; } bool all_T = false; lb ans = 0.0; if(pre_T){ lb nn = pre_T; int cur = pre_T; if(cur >= n){ all_T = true; goto TT; } if(str[cur] == 'S') ans += sqrt(Pow(nn - 0.5) + Pow(2 - sqrt(3)) / 4) + 0.5; else if(str[cur] == 'C'){ lb A = (4 * Pow(nn)) / Pow(sqrt(3) - 1.0) + 1.0; lb B = -(2 * nn) / Pow(sqrt(3) - 1); lb C = (1.0 / 4.0) / Pow(sqrt(3) - 1.0) - 1.0 / 4.0; lb delta = Pow(B) - 4 * A * C; lb x1 = (-B - sqrt(delta)) / (2 * A); lb y = sqrt(1.0 / 4.0 - Pow(x1)); lb t2 = Pow(x1) + Pow(1 / 2.0 - y); lb ct = (1/2.0 - t2) * 2; lb alf = acos(ct); lb L = alf / 2.0 + sqrt(Pow(x1 - nn) + Pow(y - (sqrt(3) - 1.0) / 2.0)); ans += L; } } if(last_T){ lb nn = last_T; int cur = n - 1 - last_T; if (cur < 0){ all_T = true; goto TT; } if (str[cur] == 'S') ans += sqrt(Pow(nn - 0.5) + Pow(2 - sqrt(3)) / 4) + 0.5; else if (str[cur] == 'C'){ lb A = (4 * Pow(nn)) / Pow(sqrt(3) - 1.0) + 1.0; lb B = -(2 * nn) / Pow(sqrt(3) - 1); lb C = (1.0 / 4.0) / Pow(sqrt(3) - 1.0) - 1.0 / 4.0; lb delta = Pow(B) - 4 * A * C; lb x1 = (-B - sqrt(delta)) / (2 * A); lb y = sqrt(1.0 / 4.0 - Pow(x1)); lb t2 = Pow(x1) + Pow(1 / 2.0 - y); lb ct = (1/2.0 - t2) * 2; lb alf = acos(ct); lb L = alf / 2.0 + sqrt(Pow(x1 - nn) + Pow(y - (sqrt(3) - 1.0) / 2.0)); ans += L; } } TT: if(all_T) ans += n - 1; else ans += n - pre_T - last_T - 1; ans += n; if (str[0] == 'S') ans += 1.5; if (str[len - 1] == 'S') ans += 1.5; if (str[0] == 'C') ans += PI / 2.0 - 0.5; if (str[len - 1] == 'C') ans += PI / 2.0 - 0.5; if (str[0] == 'T') ans += 1; if (str[len - 1] == 'T') ans += 1; printf("%.10f\n", (double)ans); } return 0; }