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 iscost i, 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
题意：
给定n个点的坐标，先问这些点能否组成一个凸包，假设是凸包。问用不相交的线来切这个凸包使得凸包仅仅由三角形组成，依据costi, j = |xi + xj| * |yi + yj| % p算切线的费用，问最少的分割费用。
思路：
先判定凸包，求凸包后看点的个数有没有变化。
然后区间dp，dp[i][j]表示切凸多边形i~j时的最小花费，特殊情况顶点个数为2或者3时不用切了为0.
如图i~j引入两条切线ik和kj将凸多边形分为两个凸多边形和一个三角形。
转移dp[i][j]=min(dp[i][k]+dp[k][j]+cost[i][k]+cost[k][j]);
cost[i][j]为切i、j两点时的花费，当j=i+1时花费为0.（凸多边形为三角形）
代码：
#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
*/
