# Sha 1425-metal (recurrence)

Source: Internet
Author: User
Ultraviolet A 1425-metal

Given a metal sheet, there are some points on it. Now there is a cutting machine that needs to cut out a monotonic quadrilateral consisting of all points and ask how many situations there are.

Idea: recursive, set DP [I] [J], I as the top vertex, and J as the bottom vertex. Now add another vertex K to come in, therefore, the original DP [I] [J] must have one dimension of k-1, and the enumeration of the other one is all. In the process of adding more points, you should also consider whether or not you can add them. Write a judgment function and use the vector cross product to judge the enumeration of all vertices in the connection line. If the above line is used, it cannot be a bit above. If it is a line below, it cannot be a bit below.

Code:

`#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 55;int t, n;struct Point {    double x, y;    void read() {scanf("%lf%lf", &x, &y);    }} p[N];long long dp[N][N];bool cmp(Point a, Point b) {    return a.x < b.x;}double xmul(Point a1, Point a2, Point b1, Point b2) {    double ax = a2.x - a1.x, ay = a2.y - a1.y;    double bx = b2.x - b1.x, by = b2.y - b1.y;    return ax * by - bx * ay;}bool judge(int s, int e, double flag) {    for (int i = s + 1; i < e; i++) {double ans = xmul(p[s], p[i], p[s], p[e]) * flag;if (ans <= 0) return false;    }    return true;}long long solve() {    memset(dp, 0, sizeof(dp));    dp[1][0] = dp[0][1] = 1;    for (int i = 2; i < n; i++) {for (int j = 0; j < i - 1; j++) {    int k = i - 1;    dp[i][j] += dp[k][j];    dp[j][i] += dp[j][k];}for (int j = 0; j < i - 1; j++) {    if (judge(j, i, 1))dp[i][i - 1] += dp[j][i - 1];    if (judge(j, i, -1))dp[i - 1][i] += dp[i - 1][j];}    }    long long ans = 0;    for (int i = 0; i < n - 1; i++) {if (judge(i, n - 1, 1))    ans += dp[i][n - 1];if (judge(i, n - 1, -1))    ans += dp[n - 1][i];    }    return ans;}int main() {    scanf("%d", &t);    while (t--) {scanf("%d", &n);for (int i = 0; i < n; i++)    p[i].read();sort(p, p + n, cmp);printf("%lld\n", solve() / 2);    }    return 0;}`

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