考慮區間dp

$dp_{i,j}$表示$i$點到$j$點所截下來的圖形的方案數，如下圖中陰影部分的劃分方案數

$dp_{i,j}=\sum\limits_^dp_{i,x}dp_{x,j},$表示先以$ij$為$1$邊剖出一個三角形，再按照乘法原理與加法原理得到結果,如下圖

轉移條件為$x$點在$ji$的右邊，如上圖,用叉乘¹判斷

\(\mathfrak{Talk\ is\ cheap,show\ you\ the\ code.}\)

#include<map>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
# define ll long long
# define read read1<int>()
# define Type template<class T>
Type T read1(){
    T t=0;
    bool vis=0;
    char k;
    do (k=getchar())=='-'&&(vis=1);while('0'>k||k>'9');
    while('0'<=k&&k<='9')t=(t<<3)+(t<<1)+(k^'0'),k=getchar();
    return vis?-t:t;
}
Type void print(T arg1){cout<<arg1;}
Type void get(T arg1){cout<<arg1;}
template<typename T,typename... Ts>
    void print(T arg1, Ts... arg_left){
        cout<<arg1<<" ";
        print(arg_left...);
    }
template<typename T,typename... Ts>
    void get(T arg1, Ts... arg_left){
        cout<<arg1<<" ";
        print(arg_left...);
    }
# define mod 1000000007
# define A pair<int,int>
A a[205];
int s,dp[205][205];
A operator -(const A &x,const A &y){return A(x.first-y.first,x.second-y.second);}
ll operator *(const A &x,const A &y){return 1ll*x.first*y.second-1ll*x.second*y.first;}
int main(){
    s=read;ll t=0;
    for(int i=1;i<=s;++i){
        a[i]=make_pair(read,read);
        if(i!=1)t+=a[i-1]*a[i];
    }
    t+=a[s]*a[1];
    for(int i=1;i<s;++i)
        dp[i][i+1]=1;
    if(t>0)reverse(a+1,a+s+1);
    for(int i=s;--i;)
        for(int j=i+2;j<=s;++j)
            for(int k=i+1;k<j;++k)
                if((a[j]-a[i])*(a[k]-a[i])>0)
                    dp[i][j]=(dp[i][j]+1ll*dp[i][k]*dp[k][j])%mod;
    printf("%d",dp[1][s]);
    return 0;
}
 

1. $\overrightarrow \times \overrightarrow>0$時$x$在$ji$右邊↩