离散化+线段树

题目网址：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=109331#problem/D

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

Sample Output

4

题意： 有一段长为1000，000，000的墙，这段墙分为单位长度的小段，然后在上面贴n份广告，可以覆盖，给出了广告纸贴的起始位置，求最后又多少广告能看见；

思路： 先将广告的始末位置数据存入结构体数组，将结构体按照id（位置）从小到大排序，然后进行离散化处理，题目给了墙壁长1000,000,000所以数据大小为1~1000,000,000 而输入只有10000张广告，所以离散化处理后，数据大小为1~20000之间，大大缩小了数据大小，便于建立线段树。接着再建立1~20000的线段树，将结构体数组按照id（贴广告的先后顺序）从大到小排序，若id相等，按照x值从小到大排序。最后在线段树中处理，计算能看见的广告数量。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)
const int N=100010;    ///注意范围,否则RE
int flag;

struct Tree
{
    int l,r;
    bool vis;   ///vis 是这块区域是否完全被覆盖
}tree[N<<2];

struct Point
{
    int id,x;
}post[N<<2];

int cmp1(Point a,Point b)
{
    return a.x<b.x;
}

int cmp2(Point a,Point b)
{   ///将结构体按照id从大到小排序，若id相等，按照x值从小到大排序；
    if(a.id==b.id)
        return a.x<b.x;
    return a.id>b.id;
}

void PushUp(int rt)
{
    tree[rt].vis=tree[L(rt)].vis && tree[R(rt)].vis;
}

void build(int L,int R,int rt)
{
    tree[rt].l=L;
    tree[rt].r=R;
    tree[rt].vis=0;
    if(tree[rt].l==tree[rt].r)
        return ;
    int mid=(L+R)>>1;
    build(L,mid,L(rt));
    build(mid+1,R,R(rt));
}

void query(int L,int R,int rt)
{
    if(tree[rt].vis)///表示这块区域已经被覆盖，直接返回；
        return ;
    if(tree[rt].l==L && tree[rt].r==R)
    {   ///找到这段贴海报的区间，标记flag，并标记为已经覆盖；
        tree[rt].vis=1;
        flag=1;
        return ;
    }
    int mid=(tree[rt].l+tree[rt].r)>>1;
    ///二分查找线段树；
    if(R<=mid)
        query(L,R,L(rt));
    else if(L>=mid+1)
        query(L,R,R(rt));
    else
    {
        query(L,mid,L(rt));
        query(mid+1,R,R(rt));
    }
    PushUp(rt);///判断并标记着这段区间是否被完全覆盖；
}

int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<2*n;i+=2)
        {
            scanf("%d%d",&post[i].x,&post[i+1].x);
            post[i].id=i;
            post[i+1].id=i;
        }
        sort(post,post+2*n,cmp1);
        int tot=0,pre=0;
        for(int i=0;i<2*n;i++)
        {     ///离散化，
            if(post[i].x==pre)
                post[i].x=tot;
            else
            {
                pre=post[i].x;
                post[i].x=++tot;
            }
        }
        build(1,2*n,1);
        sort(post,post+2*n,cmp2);     ///排序，从后往前贴
        int ans=0;
        for(int i=0;i<2*n;i+=2)
        {
            int l=post[i].x;
            int r=post[i+1].x;
            flag=0;
            query(l,r,1);
            if(flag)
            ans++;
        }
        printf("%d
",ans);
    }
    return 0;
}