Codeforces 437E The Child and Polygon(间隔DP)

题目链接：Codeforces 437E The Child and Polygon

题目大意：给出一个多边形，问说有多少种切割方法。将多边形切割为多个三角形。

解题思路：首先要理解向量叉积的性质,一開始将给出的点转换成顺时针。然后用区间dp计算。dp[i][j]表示从点i到点j能够有dp[i][j]种分割方法。然后点i和点j能否够做为分割线。要经过推断。即在i和j中选择的话点k的话，点k要在i，j的逆时针方。

#include <cstdio>
#include <cstring>
#include <algorithm>

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
", solve(0, n-1));
    return 0;
}