POJ2528 Mayor's posters(線段樹+離散化+染色)

Mayor's posters

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 68924		Accepted: 19861

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：給出一些個區間，在這些區間上貼海報，後貼的海報可能會覆蓋前面的海報，問最後能看見幾個海報；

給出的數據只有1w對，但是每個區間的數字可能會很大，用線段樹維護的話，如果不離散化的話，可能無法維護。

離散化時要注意可能出現

[1,10]   [1,5]   [8,10]這樣普通離散化的話可能會變成[1,4]   [1,2]   [3,4]導致丟失中間區間，故對於相鄰的兩個數字的話如果差值大於1，就新插入一個新的元素。

離散化用unique，重複的元素被放在數組尾部。

代碼如下：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N=100005*5;
int l[N];
int r[N];
int num[N];
int tree[N];
int has[N];
int ans=0;
int res[N];

void push(int i)
{
    if(tree[i]!=0)
    {
        tree[i*2+1]=tree[i*2]=tree[i];
        tree[i]=0;
    }
}
void build(int l,int r,int i)
{
    tree[i]=0;
    if(l==r)
        return ;
    int mid=(l+r)/2;
    build(l,mid,i*2);
    build(mid+1,r,i*2+1);
}
void updata(int l,int r,int L,int R,int i,int val)
{
    if(l>=L&&r<=R)
    {
        tree[i]=val;
        return ;
    }
    push(i);
    int mid=(l+r)/2;
    if(L<=mid)
        updata(l,mid,L,R,i*2,val);
    if(R>mid)
        updata(mid+1,r,L,R,i*2+1,val);
}
void query(int l,int r,int i)
{
    if(l==r)
    {
        if(has[tree[i]]==0&&tree[i]!=0)
        {
            ans++;
            has[tree[i]]=1;
        }
        return ;
    }
    push(i);
    int mid=(l+r)/2;
    query(l,mid,i*2);
    query(mid+1,r,i*2+1);
}
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        memset(has,0,sizeof(has));
        scanf("%d",&n);
        int tot=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d %d",&l[i],&r[i]);
            num[++tot]=l[i];
            num[++tot]=r[i];
        }
        sort(num+1,num+tot+1);
        tot=unique(num+1,num+tot+1)-num-1;
        for(int i=tot;i>=2;i--)
        {
            if(num[i]-num[i-1]>1)
                num[++tot]=num[i]-1;
        }
        sort(num+1,num+tot+1);
        build(1,tot,1);
        for(int i=1;i<=tot;i++)
            res[i]=i;
        for(int i=1;i<=n;i++)
        {
            int indexl=upper_bound(num+1,num+tot+1,l[i])-num-1;
            int indexr=upper_bound( num+1,num+tot+1,r[i])-num-1;
            int le=res[indexl];
            int ri=res[indexr];
            updata(1,tot,le,ri,1,i);
        }
        ans=0;
        query(1,tot,1);
        printf("%d\n",ans);
    }
    return 0;
}