poj2528 Mayor's posters (线段树+离散化)

Mayor's posters

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 46035		Accepted: 13332

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：往墙上贴n张海报,每张海报左端点l,右端点r,后面贴的海报会覆盖前面,问最后可看见多少张海报。

 

思路：由于坐标范围太大,要用到离散化,最后用线段数求解。

 

离散化解释：有一条1到10的数轴（长度为9），给定4个区间[2,4] [3,6] [8,10] [6,9]，覆盖关系就是后者覆盖前者，每个区间染色依次为 1 2 3 4。

现在我们抽取这4个区间的8个端点，2 4 3 6 8 10 6 9

然后删除相同的端点，这里相同的端点为6，则剩下2 4 3 6 8 10 9

对其升序排序，得2 3 4 6 8 9 10

然后建立映射

2     3     4     6     8     9    10

↓    ↓    ↓    ↓    ↓    ↓    ↓

1     2     3     4     5     6     7

那么新的4个区间为 [1,3] [2,4] [5,7] [4,6]，覆盖关系没有被改变。新数轴为1到7，即原数轴的长度从9压缩到6，显然构造[1,7]的线段树比构造[1,10]的线段树更省空间，索也更快，但是求解的结果却是一致的。

但是只是这样简单的离散化是错误的，如三张海报为：1~10 1~4 6~10

离散化时 X[ 1 ] = 1, X[ 2 ] = 4, X[ 3 ] = 6, X[ 4 ] = 10
第一张海报时：墙的1~4被染为1；
第二张海报时：墙的1~2被染为2，3~4仍为1；
第三张海报时：墙的3~4被染为3，1~2仍为2。
最终，第一张海报就显示被完全覆盖了，于是输出2，但实际上明显不是这样，正确输出为3。

新的离散方法为：在相差大于1的数间加一个数，例如在上面1 4 6 10中间加5（算法中实际上1，4之间，6，10之间都新增了数的）

X[ 1 ] = 1, X[ 2 ] = 4, X[ 3 ] = 5, X[ 4 ] = 6， X[ 5 ] = 10

这样之后，第一次是1~5被染成1；第二次1~2被染成2；第三次4~5被染成3

最终，1~2为2，3为1，4~5为3，于是输出正确结果3。

 

代码：

 

#include<cmath>
#include<cstdio>
#include<cstring>
#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;

#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , mid
#define rson rs , mid + 1 , r
#define root 1 , 1 , n
#define rt o , l , r
const int N = 20100;

struct post
{
    int L,R;
}po[N];

struct node
{
    int le,re;
    bool cover;
}s[N*8];//这里因为有离散化多加了点,所以 *8,平时 *4就好

void build (int o , int l , int r)
{
    s[o].le = l,s[o].re = r;
    s[o].cover = false;
    if (l == r)
    {
        return;
    }
    int mid = ( l + r ) >> 1;
    build(lson);
    build(rson);
}
int update(int o , int l , int r)
{
    if(s[o].cover == true)return 0;
    if(s[o].le == l && s[o].re == r)
    {
        s[o].cover = true;
        return 1;
    }
    int res,mid = ( s[o].le + s[o].re ) >> 1;
    if(r <= mid)
        res = update(ls, l, r);//这里还有下面我写成了res = update(lson);找了半天错。。。
    else if(l >= mid+1)
        res = update(rs, l, r);
    else
    {
        int a1=update(lson);
        int a2=update(rson);
        res = a1 || a2;
    }
    if(s[ls].cover && s[rs].cover)//更新节点
        s[o].cover = true;
    return res;
}

vector <int> qu;
int dui[10000005];
int main()
{
    //freopen("in.txt","r",stdin);
    int t,n,cnt;
    scanf("%d",&t);
    while(t--)
    {
        qu.clear();//清空！
        scanf("%d",&n);
        for(int i=0; i<n; i++)
        {
            scanf("%d%d",&po[i].L,&po[i].R);
            qu.push_back(po[i].L),qu.push_back(po[i].R);
        }
        sort(qu.begin(),qu.end());
        qu.resize(unique(qu.begin(),qu.end())-qu.begin());
        int ncount=qu.size();
        for(int i=0; i<ncount-1; i++)
        {
            if(qu[i+1]-qu[i]>1)qu.push_back(qu[i]+1);
        }
        sort(qu.begin(),qu.end());
        ncount=qu.size();
        //for(int i=0; i<ncount; i++)printf("qu[%d]=%d\n",i,qu[i]);

        for(int i=0; i<ncount; i++)
            dui[qu[i]]=i+1;
        build(1,1,ncount);
        cnt = 0;
        for(int i=n-1; i>=0; i--)
        {
            if(update(1,dui[po[i].L],dui[po[i].R]))
                cnt++;
        }
        printf("%d\n",cnt);
    }
    return 0;
}