題目鏈接：Codeforces 437E The Child and Polygon

題目大意：給出一個多邊形，問說有多少種分割方法，將多邊形分割為多個三角形。

解題思路：首先要理解向量叉積的性質,一開始將給出的點轉換成順時針，然後用區間dp計算。dp[i][j]表示從點i到點j可以有dp[i][j]種切割方法。然後點i和點j是否可以做為切割線，要經過判斷，即在i和j中選擇的話點k的話，點k要在i，j的逆時針方。

#include 
#include 
#include 

using namespace std;
typedef long long ll;

const int N = 205;
const ll MOD = 1e9+7;

struct point {
    ll x, y;
    ll operator * (const point& c) const {
        return x * c.y - y * c.x;
    }
    point operator - (const point& c) const {
        point u;
        u.x = x - c.x;
        u.y = y - c.y;
        return u;
    }
}p[N];
int n;
ll dp[N][N];

void init () {
    scanf("%d", &n);
    memset(dp, -1, sizeof(dp));

    for (int i = 0; i < n; i++)
        scanf("%lld %lld", &p[i].x, &p[i].y);

    ll tmp = 0;
    p[n] = p[0];
    for (int i = 0; i < n; i++)
        tmp += p[i] * p[i+1];

    if (tmp < 0)
        reverse(p, p + n);
}

ll solve (int l, int r) {
    if (dp[l][r] != -1)
        return dp[l][r];

    ll& ans = dp[l][r];
    ans = 0;

    if (r - l == 1)
        return ans = 1;

    for (int i = l + 1; i < r; i++) {
        if ((p[l] - p[r]) * (p[i] - p[r]) > 0)
            ans = (ans + solve(l, i) * solve(i, r)) % MOD;
    }
    return ans;
}

int main () {
    init();
    printf("%lld\n", solve(0, n-1));
    return 0;
}