poj1259The Picnic & hdu6219 Empty Convex Polygon（17沈阳区域赛C）最大空凸包

Posted 2021-01-12 wyboooo

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=6219

http://poj.org/problem?id=1259

一份代码A两题。

题意：

给n个点，求一个面积最大的空凸包。

思路：

空凸包，就是一个内部没有其他给定点的凸包。

详细讲解见：https://blog.csdn.net/nyroro/article/details/45268767

总的来说就是先枚举凸包的最左下角的点O。按照极坐标排序。

dp[i][j]表示组成凸包的最后一个三角形的是Oij时的最大面积。dp[i][j]=max(dp[i][j],triangle(O,i,j)+dp[j][k])

再枚举凸包上最后的一个点i，枚举所有比i小的合法的j。

具体讲解见：https://blog.csdn.net/cdsszjj/article/details/79366813

复杂度O(n^3)

  1 #include<iostream>
  2 #include<cmath>
  3 #include<algorithm>
  4 #include<stdio.h>
  5 #include<cstring>
  6 #include<vector>
  7 #include<map>
  8 #include<set>
  9 
 10 #define inf 0x3f3f3f3f
 11 using namespace std;
 12 typedef long long LL;
 13 
 14 const int maxn = 105;
 15 struct point{
 16     double x, y;
 17     point(){}
 18     point(double _x, double _y):x(_x), y(_y){}
 19     point operator + (const point &b) const{return point(x + b.x, y + b.y);}
 20     point operator - (const point &b) const{return point(x - b.x, y - b.y);}
 21     double operator * (const point &b) const {return x * b.y - y * b.x;}
 22     double len() const {return x * x + y * y;}
 23     /*int operator < (const point &a) const
 24     {
 25         if((*this)*a > 0 || (*this) *a == 0 && len() < a.len())
 26             return 1;
 27         return 0;
 28     }*/
 29 
 30 }a[maxn], p[maxn], yuan;
 31 /*bool cmp(const point &a, const point &b)
 32 {
 33     int c = a * b;
 34     if(c == 0)return a.len() < b.len();
 35     return c > 0;
 36 }*/
 37 double dp[maxn][maxn], ans;
 38 int t, n, m;
 39 bool cmp(const point &a, const point &b)
 40 {
 41     int res = (a - yuan) * (b - yuan);
 42     if(res)return res > 0;
 43     return (a - yuan).len() < (b - yuan).len();
 44 }
 45 void solve()
 46 {
 47     memset(dp, 0, sizeof(dp));
 48     sort(p + 1, p + m + 1, cmp);
 49     for(int i = 1; i <= m; i++){
 50         int j = i - 1;
 51         while(j && !((p[i] - yuan) * (p[j] - yuan)))j--;
 52         bool bz = (j == i - 1);
 53         while(j){
 54             int k = j - 1;
 55             while(k && (p[i] - p[k]) * (p[j] - p[k]) > 0)k--;
 56             double area = fabs((p[i] - yuan) * (p[j] - yuan)) / 2;
 57             if(k) area += dp[j][k];
 58             if(bz) dp[i][j] = area;
 59             ans = max(ans, area);
 60             j = k;
 61         }
 62         if(bz){
 63             for(int j = 1; j < i; j++){
 64                 dp[i][j] = max(dp[i][j], dp[i][j - 1]);
 65             }
 66         }
 67     }
 68 }
 69 
 70 int getint()
 71 {
 72     int i = 0, f = 1;
 73     char c;
 74     for(c = getchar(); (c != ‘-‘) && (c < ‘0‘ || c > ‘9‘); c = getchar());
 75     if(c == ‘-‘)f = -1, c = getchar();
 76     for(;c >= ‘0‘ && c <= ‘9‘; c = getchar())i = (i << 3) + (i << 1) + c - ‘0‘;
 77     return i * f;
 78 }
 79 
 80 int main(){
 81 
 82     //scanf("%d", &t);
 83     t = getint();
 84     while(t--){
 85         //scanf("%d", &n);
 86         n = getint();
 87         ans = 0;
 88         for(int i = 1; i <= n; i++){
 89             cin>>a[i].x>>a[i].y;
 90         }
 91         for(int i = 1; i <= n; i++){
 92             yuan = a[i];
 93             m = 0;
 94             for(int j = 1; j <= n; j++){
 95                 if(a[j].y > a[i].y || a[j].y == a[i].y && a[j].x > a[i].x)
 96                     p[++m] = a[j];//只取右上角的点
 97             }
 98             solve();
 99 
100         }
101         printf("%0.1f
", ans);
102     }
103     return 0;
104 }