poj2528-Mayor's posters (线段树+离散化)

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

贴海报,在1~1千万长度的范围内贴海报,海报没有宽度,只有长度,最多贴1万张.

 

输入案例个数c

输入海报个数n

输入海报左边界li,和右边界ri

问最后能看到几种海报

 

如果直接用线段树,必然会爆掉

所以需要离散化处理

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define maxn 10240
using namespace std;

struct node{
int l,r;
}np[maxn];

bool hash1[maxn];               //用于记录出现的广告数量
int col[maxn<<4],x[maxn<<2],ans;//col记录广告的代号,x存储广告的离散后的左边界,右边界

void pushdown(int rt)
{
    if(col[rt]!=-1)
    {
        col[rt<<1]=col[rt<<1|1]=col[rt];
        col[rt]=-1;
    }
}

void update(int L,int R,int c,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        col[rt]=c;
        return ;
    }
    pushdown(rt);
    int m=(l+r)>>1;
    if(L<=m) update(L,R,c,lson);
    if(R>m)  update(L,R,c,rson);
}

void query(int l,int r,int rt)
{
    if(col[rt]!=-1)
    {
        if(!hash1[col[rt]]) ans++;
        hash1[col[rt]]=true;
        return;
    }
    if(l==r) return;
    int m=(l+r)>>1;
    query(lson);
    query(rson);
}

int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        int m=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&np[i].l,&np[i].r);
            x[m++]=np[i].l,x[m++]=np[i].r;
        }
        sort(x,x+m);
        int k=unique(x,x+m)-x;//去重
        memset(col,-1,sizeof(col));
        for(int i=0;i<n;i++)
        {
            int l=lower_bound(x,x+k,np[i].l)-x;
            int r=lower_bound(x,x+k,np[i].r)-x;//离散化处理
            update(l,r,i,0,k,1);
        }
        memset(hash1,false,sizeof(hash1));
        ans=0;
        query(0,m,1);
        printf("%d\n",ans);
    }
    return 0;
}