poj 3109 Inner Vertices(树状数组）

Inner Vertices

Description

There is an infinite square grid. Some vertices of the grid are black and other vertices are white.

A vertex V is called inner if it is both vertical-inner and horizontal-inner. A vertex V is called horizontal-inner if there are two such black vertices in the same row that V is located between them. A vertex V is
called vertical-inner if there are two such black vertices in the same column that V is located between them.

On each step all white inner vertices became black while the other vertices preserve their colors. The process stops when all the inner vertices are black.

Write a program that calculates a number of black vertices after the process stops.

Input

The first line of the input file contains one integer number n (0 ≤ n ≤ 100 000) — number of black vertices at the beginning.

The following n lines contain two integer numbers each — the coordinates of different black vertices. The coordinates do not exceed 10⁹ by their absolute values.

Output

Output the number of black vertices when the process stops. If the process does not stop, output -1.

Sample Input

4
0 2
2 0
-2 0
0 -2

Sample Output

5

Hint

看了一天多的别人的代码：http://blog.csdn.net/z309241990/article/details/38638243，太弱了

预处理标记第i个点是最上面的还是最下面的点，如果是最下面的点且不是最下面的点，就增加一列被染色的点，

如果是最上面的点且不是最下方的点，肯定少一列被染色的点，最后以y坐标为扫描线，看这段区间有几列被染

色。

代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=200000+100;
struct node
{
    int x;
    int y;
}p[maxn];
int a[maxn];
int b[maxn];
int n,cur;
bool vis[maxn],l[maxn],r[maxn];
bool cmp(node u,node v)
{
    if(u.y==v.y)
        return u.x<v.x;
    return u.y<v.y;
}
int low(int k)
{
    return k&(-k);
}
void update(int k,int v)
{
    while(k<maxn)
    {
        a[k]+=v;
        k+=low(k);
    }
}
long long sum(int k)
{
    long long ans=0;
    while(k>0)
    {
        ans+=a[k];
        k-=low(k);
    }
    return ans;
}
void init()//预处理
{
    memset(a,0,sizeof(a));//初始化树状数组
    sort(b,b+cur);
    cur=unique(b,b+cur)-b;//排序，离散化
    for(int i=0;i<n;i++)
    {
        p[i].x=lower_bound(b,b+cur,p[i].x)-b+1;
        p[i].y=lower_bound(b,b+cur,p[i].y)-b+1;
    }
    sort(p,p+n,cmp);//按y排序，已y坐标建扫描线
    memset(vis,0,sizeof(vis));
    for(int i=0;i<n;i++)
    {
       if(!vis[p[i].x])//预处理，k位置在下方第一次出现
       {
           l[i]=1;
           vis[p[i].x]=1;
       }
       else
       l[i]=0;
    }
    memset(vis,0,sizeof(vis));
    for(int i=n-1;i>=0;i--)
    {
        if(!vis[p[i].x])//第k个点的x坐标在上方出现第一次出现
        {
            r[i]=1;
            vis[p[i].x]=1;
        }
        else
        r[i]=0;
    }
}
void solve()
{
    long long ans=0;
    for(int i=0;i<n;)
    {
        int j=i;
       while(j<n-1&&p[j].y==p[j+1].y)//同一条扫描线
       j++;
       for(int k=i;k<=j;k++)
       {
           if(!l[k]&&r[k])// 最上面的点，且下面存在被染色的点
           {
               update(p[k].x,-1);
           }
       }
       if(i!=j)
       ans+=(sum(p[j].x-1)-sum(p[i].x));//p[j].x-1到p[i].x间被改变颜色的点
       if(i<j)
       {
           for(int k=i+1;k<j;k++)
           if(!l[k]&&!r[k])//原来是黑色的点
           ans--;
       }
       while(i<=j)
       {
           if(l[i]&&!r[i])//p[i].x是最底下的位置，且上面有元素，p[i].x上的要被染色
           update(p[i].x,1);
           i++;
       }
    }
    printf("%I64d\n",ans+n);
}
int main()
{
    while(~scanf("%d",&n))
    {
        cur=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&p[i].x,&p[i].y);
            b[cur++]=p[i].x;
            b[cur++]=p[i].y;
        }
        init();
        solve();
    }
    return 0;
}