poj2528 Mayor's posters（线段树+离散化）

题目传送门

Mayor's posters

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 79846		Accepted: 22978

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

题意：给你一些海报的宽度（高度一样），现在有一面墙（10000000）

      再给你n（n<20000）个海报的宽度（l,r）

      求最后没有被完全覆盖的有多少张？

题解：我们可以以墙宽建立线段树，然后求不同数的个数即可

      但是此题的墙很长，而n很小，所以要先离散化

      一般的离散化就是sort，再unique，最后用low_bound

      但是此题用一般的离散化不行

      比如1-4 6-10 1-10 未离散化是3，普通离散化是2

      解决的办法是在两个数相距大于1是再加上一个数

      然后建树求值即可

代码：

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> PII;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define maxn 20005
int sum[maxn*4];
int lazy[maxn*8];
int has[maxn];
int ans;
struct tree{
 int l,r;
 void Input(){
 scanf("%d %d",&l,&r);
 }
}s[maxn];
void pushdown(int rt)
{
    if(lazy[rt]){
    lazy[rt<<1]=lazy[rt<<1|1]=lazy[rt];
    lazy[rt]=0;
    }
}
void update(int L,int R,int c,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        lazy[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(lazy[rt]){
        if(!has[lazy[rt]]) ans++;
        has[lazy[rt]]=1;
        return ;
    }
    if(l==r) return ;
    int m=(l+r)>>1;
    query(lson);
    query(rson);
}
int main()
{
    int T;
    scanf("%d",&T);
    int n;
    while(T--){
        scanf("%d",&n);
        memset(lazy,0,sizeof(lazy));
        memset(has,0,sizeof(has));
        ans=0;
        int tot=0;
        for(int i=0;i<n;i++)
        {
            s[i].Input();
            sum[tot++]=s[i].l;
            sum[tot++]=s[i].r;
        }
        sort(sum,sum+tot);
        tot=unique(sum,sum+tot)-sum;
        for(int i=tot-1;i>=0;i--)
        {
            if(sum[i]!=sum[i-1]+1) sum[tot++]=sum[i-1]+1;
        }
        sort(sum,sum+tot);
        /*for(int i=0;i<tot;i++)
            cout<<sum[i]<<endl;*/
        for(int i=0;i<n;i++)
        {
            int l=lower_bound(sum,sum+tot,s[i].l)-sum;
            int r=lower_bound(sum,sum+tot,s[i].r)-sum;
            update(l,r,i+1,0,tot,1);
        }
        query(0,tot,1);
        printf("%d\n",ans);
    }

    return 0;
}