zoj3537 cake 区间dp+凸包

Cake

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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 30 01 10 2

Sample Output

0

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Author: LI, Zezhou

Contest: ZOJ Monthly, September 2011

题意：给定n个点的坐标，先问这些点时候能组成一个凸包，如果是凸包，问用不相交的先来切这个凸包使得凸包只由三角形组成，根据cos[i][j]=|xi+xj |*|yi +yj | %p算切线的费用，问最少的切割费。

思路：经典的最优三角剖分模型。

  首先，先判断给定的n个点的坐标是否构成凸包；之后我们选以1和n为起始点的凸包，由于切割的线不能相交，所以选择1到n之间的任意一点s组成三角形之后，该凸包被分成三部分，一个是以1和s为起始点的凸包k0,一个是以s和n为起始点的凸包k1，以及1和n和s构成的三角形；那么我们按照这个思路继续切割k0和k1，直至所剩均为三角形。

所以设dp[i][j]为以i和j为起始点的凸包被切成一个个三角形所花的最小费用。那么状态转移方程为:dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j+cost[i][k]+cos[k][j]),其中i<k<j，记忆化搜索之。详见程序：

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;const int MAXN=300+100;const double eps=1e-10;const int inf=0x3fffffff;int n,m;int cost[MAXN][MAXN],dp[MAXN][MAXN];struct point{    double x,y;}p[MAXN],poly[MAXN];int dcmp(double x){    if(fabs(x)<eps) return 0;    return x<0 ? -1 : 1;}bool operator ==(const point &a,const point &b){    return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0 ;}bool cmp(point a,point b){    if(a.x==b.x)        return a.y<b.y;    return a.x<b.x;}double cross(point p1,point p2,point p0){    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}int Andrew(int n){    sort(p,p+n,cmp);    int top=-1;    for(int i=0;i<n;i++)    {        while(top>0 && cross(p[i],poly[top],poly[top-1])>=0)            top--;        poly[++top]=p[i];    }    int len=top;    for(int i=n-2;i>=0;i--)    {        while(top!=len && cross(p[i],poly[top],poly[top-1])>=0)            top--;        poly[++top]=p[i];    }    return top;}int get_cost(point a,point b){    return ((int)(abs(a.x+b.x)*abs(a.y+b.y)))%m;}void dfs(int i,int j){    if(j-i==1)    {        dp[i][j]=0;        return ;    }    if(dp[i][j]!=-1)        return ;    dp[i][j]=inf;    for(int k=i+1;k<j;k++)    {        dfs(i,k); dfs(k,j);        dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+cost[i][k]+cost[k][j]);    }}int main(){    //freopen("text.txt","r",stdin);    while(~scanf("%d%d",&n,&m))    {        for(int i=0;i<n;i++)            scanf("%lf%lf",&p[i].x,&p[i].y);        int top=Andrew(n);        if(top<n)        {            printf("I can't cut.\n");            continue;        }        for(int i=0;i<n;i++)            for(int j=i+2;j<n;j++)                cost[i][j]=cost[j][i]=get_cost(poly[i],poly[j]);        memset(dp,-1,sizeof(dp));        dfs(0,n-1);        printf("%d\n",dp[0][n-1]);    }    return 0;}