poj3178 Roping the Field (计算几何 + dp)

Roping the Field

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 858		Accepted: 250

Description

Farmer John is quite the nature artist: he often constructs large works of art on his farm. Today, FJ wants to construct a giant "field web". FJ's field is large convex polygon with fences along the boundary and fence posts at each of the N corners (1 <= N <= 150). To construct his field web, FJ wants to run as many ropes as possible in straight lines between pairs of non-adjacent fence posts such that no two ropes cross. 

There is one complication: FJ's field is not completely usable. Some evil aliens have created a total of G (0 <= G <= 100) grain circles in the field, all of radius R (1 <= R <= 100,000). FJ is afraid to upset the aliens, and therefore doesn't want the ropes to pass through, or even touch the very edge of a grain circle. Note that although the centers of all the circles are contained within the field, a wide radius may make it extend outside of the field, and both fences and fence posts may be within a grain circle. 

Given the locations of the fence posts and the centers of the circles, determine the maximum number of ropes that FJ can use to create his field web. 

FJ's fence pots and the circle centers all have integer coordinates X and Y each of which is in the range 0..1,000,000.

Input

Line 1: Three space-separated integers: N, G, and R 

Lines 2..N+1: Each line contains two space-separated integers that are the X,Y position of a fence post on the boundary of FJ's field. 

Lines N+2..N+G+1: Each line contains two space-separated integers that are the X,Y position of a circle's center inside FJ's field.

Output

Line 1: A single integer that is the largest number of ropes FJ can use for his artistic creation.

Sample Input

5 3 1
6 10
10 7
9 1
2 0
0 3
2 2
5 6
8 3

Sample Output

1

Hint

Explanation of the sample: 

A pentagonal field, in which all possible ropes are blocked by three grain circles, except for the rope between fenceposts 2 and 4.

题意：

在一平面上，按顺序给定 n 个点（事实上给出的点可以按顺逆时针围成一个闭合的多边形），m 个半径为 r 的圆。要给 n 个点两两连线，要求最多能连多少条线；

连线限制：

1. 连线不能穿过任意一个圆，也不能与圆相切；
2. 连线不能在中途相交，同一个点可以连多条线；
3. 相邻的两个点不能连线（第一个点和第 n 个点是相邻的）。

思路：

首先在不考虑相交的情况下预处理出哪两个点可以连线，然后dp；

dp[i][j] 表示第 i 个点到第 j 个点最多可连多少条线，dp[i][j] = max(dp[i][k] + dp[k][j] + cnc[i][j], dp[i][j])；

其中 i < k < j；cnc[i][j] 表示 i , j 能否连线，能则为1，否则为0；

代码：

#include<map>
#include<set>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<cctype>
#include<cfloat>
#include<cstdio>
#include<memory>
#include<string>
#include<vector>
#include<cassert>
#include<csignal>
#include<cstdlib>
#include<cstring>
#include<numeric>
#include<utility>
#include<iostream>
#include<algorithm>
#include<functional>
#define LL long long
#define PB push_back
#define MAXN 171
#define RPT(I,L,R) for(int I=L;I<R;++I)
#define TPR(I,R,L) for(int I=R;I>=L;--I)
using namespace std;
template<class T> bool Umx(T &A,const T &B)
{
    return B>A?A=B,1:0;
}
template<class T> bool Umn(T &A,const T &B)
{
    return B<A?A=B,1:0;
}
const int inf=~0u>>2;

int n,m,i,j,k;
LL r;
bool cnc[MAXN][MAXN];
int f[MAXN][MAXN];

struct point
{
    LL x,y;
    bool operator < (const point &T) const
    {
        return this->y<T.y || (this->y==T.y && this->x<T.x);
    }
    void read()
    {
        scanf("%I64d%I64d",&x,&y);
    }
} p[MAXN],c[MAXN];

inline LL sqr(LL X)
{
    return X*X;
}
inline LL dot(point A,point B,point O)
{
    return (A.x-O.x)*(B.x-O.x)+(A.y-O.y)*(B.y-O.y);
}
inline LL cross(point A,point B)
{
    return A.x*B.y-B.x*A.y;
}
inline LL cross(point A,point B,point O)
{
    return (A.x-O.x)*(B.y-O.y)-(B.x-O.x)*(A.y-O.y);
}
inline LL dist2(point A,point B)
{
    return sqr(A.x-B.x)+sqr(A.y-B.y);
}

struct cmp
{
    point O;
    cmp(const point &OO):O(OO) {}
    bool operator()(const point &A,const point &B)
    {
        return cross(A,B,O)>0;
    }
};

void init()
{
    memset(f,-1,sizeof f);
    scanf("%d%d%d",&n,&m,&r);
    RPT(i,0,n) p[i].read();
    RPT(i,0,m) c[i].read();
    sort(p,p+n);
    sort(p+1,p+n,cmp(p[0]));
}

bool CrsCcl(point A,point B,point O)
{
    if (sqr(r)>=min(dist2(O,A),dist2(O,B))) return true;
    if (dot(B,O,A)<0ll || dot(A,O,B)<0ll) return false;
    return (double)r*r*dist2(A,B)>=(double)cross(A,B,O)*cross(A,B,O);
}

void Deal_Cnc()
{
    memset(cnc,false,sizeof cnc);
    RPT(i,0,n)
    RPT(j,i+2,n)
    if (!(i==0 && j==n-1))
    {
        cnc[i][j]=cnc[j][i]=true;
        RPT(k,0,n)
        if (CrsCcl(p[i],p[j],c[k]))
        {
            cnc[i][j]=cnc[j][i]=false;
            break;
        }
    }
}

int DP(int L,int R)
{
    if (f[L][R]>0) return f[L][R];
    if (R-L<2) return f[L][R]=0;
    else if (R-L==2) return cnc[L][R];
    int res=0;
    RPT(i,L+1,R) Umx(res,DP(L,i)+DP(i,R));
    return f[L][R]=res+cnc[L][R];
}

int main()
{
    init();
    Deal_Cnc();
    printf("%d\n",DP(0,n-1));
    return 0;
}

 

转载于:https://www.cnblogs.com/xiepingfu/p/7287004.html