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  • UVALive 7749 Convex Contour (计算几何)

    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
    ", (double)ans);
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7454652.html
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