hdu6219 Empty Convex Polygons （最大空凸包板子

https://vjudge.net/contest/324256#problem/L

题意：给一堆点，求最大空凸包面积。

思路：枚举凸包左下角点O，dp找出以这个点为起始位置能构成的最大空凸包面积，用dp[i][j]表示以Oi和ij为凸包最后两边所构成凸包面积的最大值。

dp[i][j] = max(dp[i][j],triangle(O,i,j)+dp[j][k])

 

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
struct Point {
    int x,y;
    Point(){};
    Point(int x,int y):x(x),y(y){}
    Point operator + (const Point &b) const {
        return Point(x+b.x,y+b.y);
    }
    Point operator -(const Point &b) const {
        return Point(x-b.x,y-b.y);
    }
    int operator *(const Point &b) const {
        return x*b.y-y*b.x;
    }
    int len() const {
        return x*x+y*y;
    }
    int operator<(const Point &a) const {
        if((*this)*a>0||(*this)*a==0&&len()<a.len()){
            return 1;
        }
        return 0;
    }
}point;
int n;
Point s[122],p[122];
int dp[122][122];

int jud(int m){
    int ans,i,j,now,k,flag,s;
    memset(dp,0,sizeof dp);
    ans = 0;
    for(i=2;i<=m;i++){
        now = i-1;
        while(now>=1&&p[i]*p[now]==0) now--;
        flag = 0;
        if(now==i-1) flag = 1;
        while(now>=1){
            s = p[now]*p[i];
            k = now-1;
            while(k>=1&&(p[now]-p[i])*(p[k]-p[now])>0) k--;
            if(k>=1) s += dp[now][k];
            if(flag) dp[i][now] = s;
            ans = max(ans,s);
            now = k;
        }
        if(flag==0) continue;
        for(j=1;j<=i-1;j++) dp[i][j] = max(dp[i][j],dp[i][j-1]);
    }
    return ans;
}

int main(){
    int t,i,j,m,ans;
    cin>>t;
    while(t--){
        cin>>n;
        for(i=1;i<=n;i++) cin>>s[i].x>>s[i].y;
        ans = 0;
        for(i=1;i<=n;i++){
            m = 0;
            for(j=1;j<=n;j++){
                if(s[j].y>s[i].y||s[j].y==s[i].y&&s[j].x>=s[i].x){
                    p[++m] = s[j]-s[i];
                }
            }
            sort(p+1,p+1+m);
            ans = max(ans,jud(m));
        }
        printf("%.1f\n",ans/2.0);
    }
    return 0;
}