POJ训练计划2528_Mayor's posters(线段树/成段更新+离散化)

解题报告

地址传送门

题意：

一些海报，覆盖上去后还能看到几张。

思路：

第一道离散化的题。

离散化的意思就是区间压缩然后映射。

给你这么几个区间[1,300000],[3,5],[6,10],[4,9]

区间左右坐标排序完就是

1,3,4,5,6,9,10,300000;

1,2,3,4,5,6, 7 ,8;

我们可以把上面的区间映射成[1,8],[2,4],[5,7],[3,6];

这样就节省了很多空间。

给线段染色， lz标记颜色。

#include <map>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
struct node
{
    int x,y;
} p[10100];
int zb[20100],_hash[10100],lz[100000],ans;
void push_down(int root )
{
    if(lz[root])
    {
        lz[root*2]=lz[root*2+1]=lz[root];
        lz[root]=0;
    }
}
void update(int root,int l,int r,int ql,int qr,int v)
{
    if(ql>r||qr<l)return;
    if(ql<=l&&r<=qr)
    {
        lz[root]=v;
        return;
    }
    push_down(root);
    int mid=(l+r)/2;
    update(root*2,l,mid,ql,qr,v);
    update(root*2+1,mid+1,r,ql,qr,v);
}
void _q(int root,int l,int r)
{
    if(lz[root])
    {
        if(!_hash[lz[root]])
            ans++;
        _hash[lz[root]]=1;
        return ;
    }
    if(l==r)return;
    int mid=(l+r)/2;
    _q(root*2,l,mid);
    _q(root*2+1,mid+1,r);
}
int main()
{
    int t,i,j,n;
    scanf("%d",&t);
    while(t--)
    {
        ans=0;
        memset(_hash,0,sizeof(_hash));
        scanf("%d",&n);
        for(i=0; i<n; i++)
        {
            scanf("%d%d",&p[i].x,&p[i].y);
            zb[i]=p[i].x;
            zb[i+n]=p[i].y;
        }
        sort(zb,zb+n*2);
        int m=unique(zb,zb+n*2)-zb;
        for(i=0; i<n; i++)
        {
            int ql=lower_bound(zb,zb+m,p[i].x)-zb+1;
            int qr=lower_bound(zb,zb+m,p[i].y)-zb+1;
            update(1,1,m,ql,qr,i+1);
        }
        _q(1,1,m);
        printf("%d\n",ans);
    }
}

Mayor‘s posters

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

POJ训练计划2528_Mayor's posters(线段树/成段更新+离散化)