poj 3109 Inner Vertices(树状数组）

Inner Vertices

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 1692		Accepted: 446
Case Time Limit: 2000MS

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 
#include 
#include 
#include 
#include 
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.x0)
    {
        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=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