Codeforces 437E The Child and Polygon

http://codeforces.com/problemset/problem/437/E

题意：求一个多边形划分成三角形的方案数

思路:区间dp，每次转移只从一个方向转移(L,R连线的某一侧)，能保证正确。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#define ll long long
const ll Mod=;
ll f[][];
struct Point{
    double x,y;
    Point(){}
    Point(double x0,double y0):x(x0),y(y0){}
}p[];
Point operator -(Point p1,Point p2){
    return Point(p1.x-p2.x,p1.y-p2.y);
}
double operator *(Point p1,Point p2){
    return p1.x*p2.y-p1.y*p2.x;
}
int read(){
    int t=,f=;char ch=getchar();
    while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
    while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
    return t*f;
}
ll solve(int l,int r){
    if (f[l][r]!=-) return f[l][r];
    if (l>r) return f[l][r]=;
    if (l+==r) return f[l][r]=;
    f[l][r]=;
    for (int i=l+;i<r;i++){
        if ((p[r]-p[l])*(p[i]-p[l])<)
         (f[l][r]+=(solve(l,i)*solve(i,r))%Mod)%=Mod;
    }
    return f[l][r];
}
int main(){
    int n=read();
    for (int i=;i<=n;i++) p[i].x=read(),p[i].y=read();
    for (int i=;i<=n;i++)
     for (int j=;j<=n;j++)
      f[i][j]=-;
    double res=;
    p[n+]=p[];
    for (int i=;i<=n;i++)
     res+=p[i]*p[i+];
    if (res<){
     for (int i=;i<=n/;i++) std::swap(p[i],p[n-i+]);
    }
    printf("%lld\n",solve(,n));
}