E - 成段更新 Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

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

Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:65536KB     64b題目大意：給你一個無限長的板子，然後依次往上面貼n張等高的海報，問你最後能看到多少張海報。

思路分析：線段樹區間更新問題，但是要注意，給的長度的可能非常大，有1e9，不加處理直接維護一個線段樹肯定會

MLE,TLE,但是我們注意到一共最多隻有2e4個點，因此我們可以用離散化的思想先對區間進行預處理，所謂的離散化，

在我理解看來就是將一個很大的區間對映為一個很小的區間，而不改變原有的大小覆蓋關係，但是注意簡單的離散化可能

會出現錯誤，給出下面兩個簡單的例子應該能體現普通離散化的缺陷:
例子一:1-10 1-4 5-10
例子二:1-10 1-4 6-10
普通離散化後都變成了[1,4][1,2][3,4]
線段2覆蓋了[1,2],線段3覆蓋了[3,4],那麼線段1是否被完全覆蓋掉了呢?
例子一是完全被覆蓋掉了,而例子二沒有被覆蓋

解決的辦法則是對於距離大於1的兩相鄰點，中間再插入一個點，本題還用到了Lazy標記的思想

直接更新區間進行標記而先不對子節點進行處理，如果需要往下更新再將標記下傳一層。

程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn=20000+100;//點的數目
int tree[maxn<<4];
int li[maxn],ri[maxn];
int lisan[3*maxn];
bool visit[3*maxn];
void pushdown(int p)
{
    tree[p<<1]=tree[(p<<1)|1]=tree[p];
    tree[p]=-1;
}
void update(int p,int l,int r,int x,int y,int v)
{
    if(x<=l&&y>=r)
    {
        tree[p]=v;
        return;
    }
    if(tree[p]!=-1) pushdown(p);
    int mid=(l+r)>>1;
    if(y<=mid) update(p<<1,l,mid,x,y,v);
    else if(x>mid) update((p<<1)|1,mid+1,r,x,y,v);
    else update(p<<1,l,mid,x,mid,v),update((p<<1)|1,mid+1,r,mid+1,y,v);
}
int ans;
void query(int p,int l,int r)
{
    //cout<<p<<endl;
    if(tree[p]!=-1)
    {
        if(!visit[tree[p]])
        {
            ans++;
            visit[tree[p]]=true;
        }
        return;
    }
    if(l==r) return;
    //if(tree[p]!=-1) pushdown(p);
    int mid=(l+r)>>1;
    query(p<<1,l,mid);
    query((p<<1)|1,mid+1,r);
}
int main()
{
    int T;
    scanf("%d",&T);
    int n;
    while(T--)
    {
        scanf("%d",&n);
        memset(tree,-1,sizeof(tree));
        memset(visit,false,sizeof(visit));
        int tot=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&li[i],&ri[i]);
            lisan[tot++]=li[i];
            lisan[tot++]=ri[i];
        }
        sort(lisan,lisan+tot);//tot是陣列長度
        int m=unique(lisan,lisan+tot)-lisan;
        int t=m;
        for(int i=1;i<t;i++)
        {
            if(lisan[i]-lisan[i-1]>1)
                lisan[m++]=lisan[i-1]+1;
        }
        sort(lisan,lisan+m);
        //for(int i=0;i<m;i++)
            //cout<<lisan[i]<<" ";
        //cout<<endl;
        for(int i=0;i<n;i++)
        {
            int x=lower_bound(lisan,lisan+m,li[i])-lisan;
            int y=lower_bound(lisan,lisan+m,ri[i])-lisan;
            update(1,0,m-1,x,y,i);
        }
        ans=0;
        query(1,0,m-1);
        printf("%d\n",ans);
    }
}