zoj3537（凸包，最优三角形，区间dp）

Cake Time Limit: 1 Second Memory Limit: 32768 KB

You want to hold a party. Here's a polygon-shaped cake on the table. You'd like to cut the cake into several triangle-shaped parts for the invited comers. You have a knife to cut. The trace of each cut is a line segment, whose two endpoints are two vertices of the polygon. Within the polygon, any two cuts ought to be disjoint. Of course, the situation that only the endpoints of two segments intersect is allowed.

The cake's considered as a coordinate system. You have known the coordinates of vexteces. Each cut has a cost related to the coordinate of the vertex, whose formula is cost_{i, j} = |x_i + x_j| * |y_i + y_j| % p. You want to calculate the minimum cost.

NOTICE: input assures that NO three adjacent vertices on the polygon-shaped cake are in a line. And the cake is not always a convex.

Input

There're multiple cases. There's a blank line between two cases. The first line of each case contains two integers, N and p (3 ≤ N, p ≤ 300), indicating the number of vertices. Each line of the following N lines contains two integers, x and y (-10000 ≤ x, y ≤ 10000), indicating the coordinate of a vertex. You have known that no two vertices are in the same coordinate.

Output

If the cake is not convex polygon-shaped, output "I can't cut.". Otherwise, output the minimum cost.

Sample Input

3 3
0 0
1 1
0 2

Sample Output

0

题意：求凸n多边形分成n-2个三角形的最小费用

分析：先求凸包，然后用最优三角形求解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))
#define sqr(a) ((a) * (a))
#define dis(a, b) sqrt(sqr(a.x - b.x) + sqr(a.y - b.y))

int n,m;
int f[305][305],dp[305][305];

struct Point
{
    int x,y;
};
Point s[305],c[305];

int mult(Point p1,Point p2,Point p0)//叉积
{

    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}

bool cmp(const Point& a,const Point &b)
{

    if(a.y == b.y)return a.x < b.x;
    return a.y < b.y;
}

int Graham(Point *s, int n)//凸包
{
    int i;
    sort(s, s+n, cmp);
    c[0] = s[0];
    c[1] = s[1];
    int top = 1;
    for(i = 0; i < n; i++)
    {
        while(top && mult(c[top],s[i],c[top-1]) >= 0)top--;
        c[++top] = s[i];
    }
    int mid = top;
    for(i = n - 2; i >= 0; i--)
    {
        while(top>mid && mult(c[top],s[i],c[top-1])>=0)top--;
        c[++top]=s[i];
    }
    return top;
}

int calc(Point a, Point b)//费用
{
    return (abs(a.x + b.x) * abs(a.y+b.y)) % m;
}

int main ()
{
    while (scanf ("%d%d",&n,&m)==2)
    {
        for (int i=0; i<n; i++)
            scanf ("%d%d",&s[i].x,&s[i].y);
        if (n==3)
        {
            cout<<"0"<<endl;
            continue;
        }
        if (Graham(s, n)<n)//求凸包
            cout<<"I can't cut."<<endl;
        else
        {
            CL(f);
            for (int i=0; i<n; i++)
            {
                for (int j=i+2; j<n; j++)
                    f[i][j] = f[j][i] = calc(c[i], c[j]);
            }
            for (int i=0; i<n; i++)
            {
                for (int j=0; j<n; j++)
                    dp[i][j] = INF;
                dp[i][(i+1)%n]=0;
            }
            for (int i=n-3; i>=0; i--)//区间dp求解
            {
                for (int j=i+2; j<n; j++)
                {
                    for (int k=i+1; k<j; k++)
                    {
                        dp[i][j]=min(dp[i][j], dp[i][k]+dp[k][j]+f[i][k]+f[k][j]);
                    }
                }
            }
            cout<<dp[0][n-1]<<endl;
        }
    }
    return 0;
}