spoj Minimax Triangulation

题解：

dp+计算几何

F[i][j]表示第i-j条边的答案

然后转移一下

代码：

#include<bits/stdc++.h>
using namespace std;
struct note{int x,y;}a[55]; 
int T,n,f[55][55];
int xot(note a,note b){return a.x*b.y-b.x*a.y;}
int cal(note a,note b,note c)
{
    return abs(xot(a,b)+xot(b,c)+xot(c,a));
}
int area(int l,int r)
{
    int s=xot(a[r],a[l]);
    for (int i=l;i<r;i++)s+=xot(a[i],a[i+1]);
    return abs(s);
}
int check(int l,int m,int r,int s)
{
    return s+area(l,m)+area(m,r)==area(l,r);
}
int deal(int l,int r)
{
    if (f[l][r]>=0) return f[l][r];
    if (l==r||l+1==r) return f[l][r]=0;
    int res=200000000;
    if (l+2==r)return f[l][r]=cal(a[l],a[l+1],a[r]);
    for (int i=l+1;i<r;i++)
     {
        int s=cal(a[l],a[i],a[r]);
        if (!check(l,i,r,s)) continue;
        int z=max(deal(l,i),deal(i,r));
        z=max(z,s);res=min(res,z);
     }
    return f[l][r]=res;
}
int main()
{
    scanf("%d",&T);
    while (T--)
     {
        scanf("%d",&n);
        memset(f,255,sizeof(f));
        for (int i=1;i<=n;i++)scanf("%d%d",&a[i].x,&a[i].y);
        printf("%.1lf\n",deal(1,n)/2.0);
     }
    return 0;
}