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
costi, j = |xi + xj| * |yi + yj| % 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
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 105
#define MAXN 100005
#define mod 1000000000
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-8
typedef long long ll;
using namespace std; int cmp(int x)
{
if(fabs(x)<eps) return 0;
if(x>0) return 1;
return -1;
}
int sqr(int x)
{
return x*x;
}
struct point
{
int x,y;
point(){};
point(int a,int b):x(a),y(b){};
void input()
{
scanf("%d%d",&x,&y);
}
friend point operator +(const point &a,const point &b)
{
return point(a.x+b.x,a.y+b.y);
}
friend point operator -(const point &a,const point &b)
{
return point(a.x-b.x,a.y-b.y);
}
friend bool operator ==(const point &a,const point &b)
{
return cmp(a.x-b.x)==0&&cmp(a.y-b.y)==0;
}
friend point operator *(const point &a,const int &b)
{
return point(a.x*b,a.y*b);
}
friend point operator *(const int &a,const point &b)
{
return point(a*b.x,a*b.y);
}
friend point operator /(const point &a,const int &b)
{
return point(a.x/b,a.y/b);
}
int norm()
{
return sqrt(sqr(x)+sqr(y));
}
};
int det(const point &a,const point &b)
{
return a.x*b.y-a.y*b.x;
}
int dot(const point&a,const point &b)
{
return a.x*b.x+a.y*b.y;
}
int dist(const point &a,const point &b)
{
return (a-b).norm();
} struct polygon_convex
{
vector<point>p;
polygon_convex(int Size=0)
{
p.resize(Size);
}
};
bool comp_less(const point &a,const point &b)
{
return cmp(a.x-b.x)<0||cmp(a.x-b.x)==0&&cmp(a.y-b.y)<0;
}
polygon_convex convex_hull(vector<point> a)
{
polygon_convex res(2*a.size()+5);
sort(a.begin(),a.end(),comp_less);
a.erase(unique(a.begin(),a.end()),a.end());
int m=0;
for(int i=0;i<a.size();i++)
{
while(m>1&&cmp(det(res.p[m-1]-res.p[m-2],a[i]-res.p[m-2]))<=0) m--;
res.p[m++]=a[i];
}
int k=m;
for(int i=int(a.size())-2;i>=0;--i)
{
while(m>k&&cmp(det(res.p[m-1]-res.p[m-2],a[i]-res.p[m-2]))<=0) m--;
res.p[m++]=a[i];
}
res.p.resize(m);
if(a.size()>1) res.p.resize(m-1);
return res;
} int n,m,ans;
int dp[305][305],cost[305][305];
vector<point> pp; int main()
{
int i,j,t;
while(~scanf("%d%d",&n,&m))
{
vector<point> pp;
point tmp;
for(i=1;i<=n;i++)
{
scanf("%d%d",&tmp.x,&tmp.y);
pp.push_back(tmp);
}
polygon_convex tb=convex_hull(pp);
if(tb.p.size()!=n) printf("I can't cut.\n");
else
{
if(n==3)
{
printf("0\n");
continue ;
}
memset(cost,0,sizeof(cost));
for(i=0;i<n;i++)
{
for(j=i+2;j<n;j++)
{
cost[i][j]=(abs(tb.p[i].x+tb.p[j].x)*abs(tb.p[i].y+tb.p[j].y))%m;
}
}
memset(dp,0x3f,sizeof(dp));
for(i=0;i<n-2;i++)
{
dp[i][i+1]=0;
dp[i][i+2]=0;
}
dp[n-2][n-1]=0;
for(int len=4;len<=n;len++)
{
for(i=0;i<n;i++)
{
j=i+len-1;
if(j>=n) break ;
for(int k=i+1;k<=j-1;k++)
{
dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+cost[i][k]+cost[k][j]);
}
}
}
printf("%d\n",dp[0][n-1]);
}
}
return 0;
}
/*
3 3
0 0
1 1
0 2 4 10
0 0
2 0
0 2
2 2 5 11
1 1
1 3
3 1
4 2
3 4
*/
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