UVALive 4221 Walk in the Park 扫描线

Walk in the Park

题目连接：

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2222

Descriptionww.co

You are responsible for inspecting the trees located in a park, to make sure they remain healthy. The location of each tree is given to you as a point in the twodimensional plane, distinct from that of every other tree. Due to recentlyreplanted grass, you are only allowed to walk through the park along a collection of paths. Each path is described by an infinite-length horizontal or vertical line in the two-dimensional plane. No tree lies on any path.

You are concerned that it may not be possible to view all the trees in the park from the paths. In particular, a tree is visible only if you can view it by standing on some path while facing in a direction perpendicular to that path; there must be no intervening tree that obstructs your view. Given the geometrical configuration of the park, please report the number of visible trees.

Input

There will be multiple input sets. For each input set, the first line will contain two integers, N and M , ( 0 < N, M$ \le$100000 ), separated by a space. N is the number of trees, and M is the number of paths.

The next N lines each contain two space-separated integers, X and Y , specifying the coordinates of a tree. X and Y may be any 32-bit integers.

The next M lines each describe a path (a vertical or horizontal line). They have the form x = K or y = K , with no spaces. K may be any 32-bit integer. x and y will be lower case.

End of the input is signified by a line with two space-separated 0's.

Output

For each input set, print a single line containing one integer, specifying the number of visible trees. There should be no blank lines between outputs.

Sample Input

6 3

-1 3

4 2

6 2

6 3

6 4

4 3

x=0

y=-1

y=5

1 2

2 3

x=5

y=5

0 0

Sample Output

5

1

Hint

题意

在二维平面上有n棵树，然后有m条道路

你能看见这棵树的定义是，这棵树存在一条到垂直于道路的线上没有任何其他的树

然后问你最多能看到多少棵树

题解：

扫描线

我们分成四个步骤跑就好了

我们从x轴正半轴看过去能看到多少棵，负半轴看过去，能看到多少，从y轴正半轴看过去，负半轴看过去

每次我们用set来记录一下vis就好了

直接暴力扫一遍

代码

#include<bits/stdc++.h>
using namespace std;

const int maxn = 100005;
inline long long read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m;
struct node
{
    int x,y;
    int flag;
    int id;
};
node p[maxn];
node l[maxn];
set<int> S;
set<int> Ans;
bool cmp1(node a,node b)
{
    return a.y<b.y;
}
bool cmp2(node a,node b)
{
    return a.y>b.y;
}
bool cmp3(node a,node b)
{
    return a.x<b.x;
}
bool cmp4(node a,node b)
{
    return a.x>b.x;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)break;
        memset(p,0,sizeof(p));
        memset(l,0,sizeof(l));
        for(int i=1;i<=n;i++)
        {
            int x,y;
            scanf("%d%d",&p[i].x,&p[i].y);
            p[i].flag = 0;
            p[i].id=i;
        }
        S.clear();
        Ans.clear();
        getchar();
        for(int i=1;i<=m;i++)
        {
            char c = getchar();
            getchar();
            int p = read();
            if(c=='x')
            {
                l[i].x=p;
                l[i].flag=2;
            }
            else
            {
                l[i].y=p;
                l[i].flag=1;
            }
        }

        vector<node> V;
        int first;

        first = 1;
        S.clear();
        V.clear();
        for(int i=1;i<=n;i++)
            V.push_back(p[i]);
        for(int i=1;i<=n;i++)
            if(l[i].flag==1)
                V.push_back(l[i]);
        sort(V.begin(),V.end(),cmp1);
        for(int i=0;i<V.size();i++)
        {
            if(V[i].flag==1)
            {
                S.clear();
                first = 0;
            }
            else
            {
                if(first==1)continue;
                if(S.count(V[i].x))
                    continue;
                S.insert(V[i].x);
                Ans.insert(V[i].id);
            }
        }

        S.clear();
        V.clear();
        first = 1;
        for(int i=1;i<=n;i++)
            V.push_back(p[i]);
        for(int i=1;i<=n;i++)
            if(l[i].flag==1)
                V.push_back(l[i]);
        sort(V.begin(),V.end(),cmp2);
        for(int i=0;i<V.size();i++)
        {
            if(V[i].flag==1)
            {
                S.clear();
                first = 0;
            }
            else
            {
                if(first==1)continue;
                if(S.count(V[i].x))
                    continue;
                S.insert(V[i].x);
                Ans.insert(V[i].id);
            }
        }

        S.clear();
        V.clear();
        first = 1;
        for(int i=1;i<=n;i++)
            V.push_back(p[i]);
        for(int i=1;i<=n;i++)
            if(l[i].flag==2)
                V.push_back(l[i]);
        sort(V.begin(),V.end(),cmp3);
        for(int i=0;i<V.size();i++)
        {
            if(V[i].flag==2)
            {
                first = 0;
                S.clear();
            }
            else
            {
                if(first==1)continue;
                if(S.count(V[i].y))
                    continue;
                S.insert(V[i].y);
                Ans.insert(V[i].id);
            }
        }

        first = 1;
        S.clear();
        V.clear();
        for(int i=1;i<=n;i++)
            V.push_back(p[i]);
        for(int i=1;i<=n;i++)
            if(l[i].flag==2)
                V.push_back(l[i]);
        sort(V.begin(),V.end(),cmp4);
        for(int i=0;i<V.size();i++)
        {
            if(V[i].flag==2)
            {
                first=0;
                S.clear();
            }
            else
            {
                if(first==1)continue;
                if(S.count(V[i].y))
                    continue;
                S.insert(V[i].y);
                Ans.insert(V[i].id);
            }
        }

        printf("%d\n",Ans.size());
    }
}