POJ 2528 Mayor's posters（線段樹+離散化）

POJ 2528 Mayor's posters（線段樹+離散化）

Language: Default Mayor's posters Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 43702 Accepted: 12763

Description

The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. The city council has finally decided to build an electoral wall for placing the posters and introduce the following rules:
Every candidate can place exactly one poster on the wall.
All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown).
The wall is divided into segments and the width of each segment is one byte.
Each poster must completely cover a contiguous number of wall segments.
They have built a wall 10000000 bytes long (such that there is enough place for all candidates). When the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. Moreover, the candidates started placing their posters on wall segments already occupied by other posters. Everyone in Bytetown was curious whose posters will be visible (entirely or in part) on the last day before elections.
Your task is to find the number of visible posters when all the posters are placed given the information about posters' size, their place and order of placement on the electoral wall.

Input

The first line of input contains a number c giving the number of cases that follow. The first line of data for a single case contains number 1 <= n <= 10000. The subsequent n lines describe the posters in the order in which they were placed. The i-th line among the n lines contains two integer numbers l_i and ri which are the number of the wall segment occupied by the left end and the right end of the i-th poster, respectively. We know that for each 1 <= i <= n, 1 <= l_i <= ri <= 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered l_i, l_i+1 ,... , ri.

Output

For each input data set print the number of visible posters after all the posters are placed.

The picture below illustrates the case of the sample input.

Sample Input

1
5
1 4
2 6
8 10
3 4
7 10

Sample Output

4

Source

Alberta Collegiate Programming Contest 2003.10.18

題意：貼海報，後面的海報可以覆蓋前面的海報，問你貼完後可以看到的海報數目（請注意是可以看到的海報）

思路：

先離散化，然後從後向前查詢，每查詢一下，就把這個區間的海報標記被覆蓋（這裡可以lazy，但是記得pushdown()）

在這裡我貼出兩份代碼，第二份我bug了一些時間

第一份

#include
#include
#include
#include

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
using namespace std;
#define N 100100

int a[N*2];
int h[10000005];

struct stud{
int le,ri;
}post[N];

struct stud1{
int le,ri;
int cover;
}f[N*4];


int cmp(int a,int b)
{
    return a=ri)
         result=query(L(pos),le,ri);
    else
        if(mid=0;i--)
            if(query(1,h[post[i].le],h[post[i].ri]))
                {
                    ans++;
                }

        printf("%d\n",ans);
    }
    return 0;
}

第二份：

#include
#include
#include
#include

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
using namespace std;
#define N 100100

int ha[10000005];
int n,ans;
int a[N*2];

struct stud{
int le,ri;
}p[N];

struct studd{
int le,ri;
int cover;
}f[N*4];

void pushdown(int pos)
{
      if(f[pos].cover)
		f[L(pos)].cover=f[R(pos)].cover=f[pos].cover;
}

void build(int pos,int le,int ri)
{
	f[pos].le=le;
	f[pos].ri=ri;
	f[pos].cover=0;
	if(le==ri) return ;

	int mid=MID(le,ri);
	build(L(pos),le,mid);
	build(R(pos),mid+1,ri);
}

int query(int pos,int le,int ri)
{
    int result=0;


   // if(f[pos].cover)
       // return 0;      因為第二份沒有這個，所以需要pushdown(pos);

	if(f[pos].le==le&&f[pos].ri==ri)
	{
		if(f[pos].cover==0)
	       result=1;
		else
			result=0;

		f[pos].cover=1;

		return result;
	}

	int mid=MID(f[pos].le,f[pos].ri);

     pushdown(pos);

     if(mid>=ri)
		result=query(L(pos),le,ri);
     else
		if(mid=0;i--)
			if(query(1,ha[p[i].le],ha[p[i].ri]))
			  ans++;

		printf("%d\n",ans);
	}
	  return 0;
}