Mayor's posters

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 60691	Accepted: 17565

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

題目大意：

題意:在牆上貼海報,海報可以互相覆蓋,問最後可以看見幾張海報
思路:這題資料範圍很大,直接搞超時+超記憶體,需要離散化:
離散化簡單的來說就是隻取我們需要的值來用,比如說區間[1000,2000],[1990,2012] 我們用不到[-∞,999][1001,1989][1991,1999][2001,2011][2013,+∞]這些值,所以我只需要1000,1990,2000,2012就夠了,將其分別對映到0,1,2,3,在於複雜度就大大的降下來了
所以離散化要儲存所有需要用到的值,排序後,分別對映到1~n,這樣複雜度就會小很多很多
而這題的難點在於每個數字其實表示的是一個單位長度(並且一個點),這樣普通的離散化會造成許多錯誤(包括我以前的程式碼,poj這題資料奇弱)
給出下面兩個簡單的例子應該能體現普通離散化的缺陷:
1-10 1-4 5-10
1-10 1-4 6-10
為了解決這種缺陷,我們可以在排序後的陣列上加些處理,比如說[1,2,6,10]
如果相鄰數字間距大於1的話,在其中加上任意一個數字,比如加成[1,2,3,6,7,10],然後再做線段樹就好了.
線段樹功能:update:成段替換 query:簡單hash

AC程式碼：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;

const int M=112345;
int L[M],R[M];
int S[M<<2];
bool Hash[M];
int col[M<<2];
int cnt;
void pushdown(int root)
{
    if(col[root]!=-1)
    {
        col[root<<1]=col[root<<1|1]=col[root];
        col[root]=-1;
    }
}
void updata(int b,int e,int c,int root,int l,int r)//b e為要查詢的區間範圍，l r為root的區間範圍
{
    if(b<=l&&r<=e)
    {
        col[root]=c;
        return;
    }
    pushdown(root);
    int mid=(l+r)/2;
    if(mid>=e)
        updata(b,e,c,root<<1,l,mid);
    else if(mid<b)
        updata(b,e,c,root<<1|1,mid+1,r);
    else
    {
        updata(b,mid,c,root<<1,l,mid);
        updata(mid+1,e,c,root<<1|1,mid+1,r);
    }

}
void query(int l,int r,int root)
{
    if(col[root]!=-1)
    {
        if(!Hash[col[root]])cnt++;
        Hash[col[root]]=true;
        return;
    }
    if(l==r)return ;
    int mid=(l+r)/2;
    query(l,mid,root<<1);
    query(mid+1,r,root<<1|1);
}
int get(int k,int n)//獲取離散化後的下標
{
    return lower_bound(S,S+n,k)-S;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        int k=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d",&L[i],&R[i]);
            S[k++]=L[i];
            S[k++]=R[i];
        }
        sort(S,S+k);
       // for(int i=0; i<k; i++)printf("%d ",S[i]);printf("\n");
        int kk=1;
        for(int i=1; i<k; i++)
        {
            if(S[i]!=S[i-1])
                S[kk++]=S[i];
        }
        for(int i=kk-1; i>0; i--)
        {
            if(S[i]!=S[i-1]+1)S[kk++]=S[i-1]+1;
        }
        sort(S,S+kk);
        //for(int i=0; i<kk; i++)printf("%d ",S[i]);printf("\n");
        memset(col,-1,sizeof(col));
        for(int i=1; i<=n; i++)
        {
            int l=get(L[i],kk);
            int r=get(R[i],kk);
           // printf("l=%d r=%d\n",l,r);
            updata(l,r,i,1,0,kk);
        }
        cnt=0;
        memset(Hash,false,sizeof(Hash));
        query(0,kk,1);
        printf("%d\n",cnt);
    }
}