codeforces 437E 区间DP

codeforces 437E

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题意：

$给定一个简单多边形各个顶点的坐标，第i个顶点的坐标为(x_i,y_i)。$
$问将其分割为n个三角形的方案数。$

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题解：

$dp[l][r]表示多边形第l个顶点到第r个顶点的区间内，分割三角形的方案数。$
$首先通过结构体重载运算符，定义向量叉乘运算，将给定的点在平面内按顺时针方向排列。$

• $枚举分割点k，若点k在点i到点j的逆时针方向上，dp[l][r] = (dp[l][r]+dp[l][k]*dp[k][r])\%mod$

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#include <bits\stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
const int N = 201;
ll dp[N][N];

struct point{
    ll x,y;
    ll operator * (const point& b) const {
        return x*b.y-y*b.x;
    }
    point operator - (const point& b) const {
        point p;
        p.x = x-b.x;
        p.y = y-b.y;
        return p;
    }
}a[N];

int main() {
    int n;
    cin >> n;
    ll s = 0;
    for(int i = 1 ; i <= n ; i++){
        cin >> a[i].x >> a[i].y;
        s += a[i-1]*a[i];
    }
    s += a[n]*a[1];
    if(s > 0){
        for(int i = 1 ; i <= n/2 ; i++){
            swap(a[i], a[n-i+1]);
        }
    }
    for(int i = 1 ; i < n ; i++){
        dp[i][i+1] = 1;
    }
    for(int len = 3 ; len <= n ; len++){
        for(int l = 1, r = len ; r <= n ; l++, r++){
            for(int k = l ; k <= r ; k++){
                if((a[r]-a[l])*(a[k]-a[l]) > 0){
                    dp[l][r] = (dp[l][r]+dp[l][k]*dp[k][r])%mod;
                }
            }
        }
    }
    cout << dp[1][n] << endl; 
    return 0; 
}