poj 2528 Mayor's posters （线段树+区间离散）

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

这道题也做了好久。首先，每个数可以代表一个格子，求最后没有被覆盖的格子数，同一段区间的格子算一个。把区间离散化，但有一个问题，会导致错误。

如[1,10]、[1 4]、[6 10],离散化后是{1,4,6,10}对应下表是{0,1,2,3} ，区间变为[0,3][0,1][2,3],第一段区间被完全覆盖了，但原区间中第一段不会完全覆盖。所以，可以在相邻两个差值大于1的数中间添加一个数，使它们不会出错。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
using namespace std;
#define N 10005
#define ll long long
struct node
{
    int x,id;
}a[N*4];      //2*n个端点+（2*n-1）
int pos[N][2],ans;
struct st
{
    int l,r,x;  //x记录颜色，初始化-1
}f[N*20];
int mark[N*4];
bool cmp(node a,node b)
{
    return a.x<b.x;
}
void creat(int t,int l,int r)
{
    f[t].l=l;
    f[t].r=r;
    f[t].x=-1;
    if(l==r)
        return ;
    int tmp=t<<1,mid=(l+r)>>1;
    creat(tmp,l,mid);
    creat(tmp|1,mid+1,r);
}

void update(int t,int l,int r,int x)
{
    if(l==f[t].l&&r==f[t].r)
    {
        f[t].x=x;
        return ;
    }
    int tmp=t<<1,mid=(f[t].l+f[t].r)>>1;
    if(f[t].x!=-1)
    {
        f[tmp].x=f[tmp|1].x=f[t].x;
        f[t].x=-1;
    }
    if(mid>=r)
        update(tmp,l,r,x);
    else if(l>mid)
        update(tmp|1,l,r,x);
    else
    {
        update(tmp,l,mid,x);
        update(tmp|1,mid+1,r,x);
    }
}

void query(int t,int l,int r)
{
    int tmp=t<<1,mid=(f[t].l+f[t].r)>>1;
    if(l==r)
    {
        if(!mark[f[t].x]&&f[t].x!=-1)  //此处注意
        {
            mark[f[t].x]=1;
            ans++;
        }
        return ;
    }
    if(f[t].x!=-1)
    {
        f[tmp].x=f[tmp|1].x=f[t].x;
        f[t].x=-1;
    }
    if(r<=mid)
        query(tmp,l,r);
    else if(l>mid)
        query(tmp|1,l,r);
    else
    {
        query(tmp,l,mid);
        query(tmp|1,mid+1,r);
    }
}
int main()
{
    int i,T,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&pos[i][0],&pos[i][1]);//左右边界
            a[i*2].x=pos[i][0];
            a[i*2].id=-(i+1);
            a[i*2+1].x=pos[i][1];
            a[i*2+1].id=i+1;
        }
        sort(a,a+n*2,cmp);
        int m=n*2;
        for(i=m-1;i>0;i--)   //添加数字，
        {
            if(a[i].x-a[i-1].x>1)
            {
                a[m].x=a[i].x-1;
                a[m++].id=0;  //ID置为0和原区间区别
            }
        }
        sort(a,a+m,cmp);
        int t=1,x=a[0].x;  //t为区间重新编号
        for(i=0;i<m;i++)
        {
            if(x!=a[i].x)
            {
                t++;
                x=a[i].x;
            }
            if(a[i].id<0)
                pos[-a[i].id-1][0]=t;
            else if(a[i].id>0)
                pos[a[i].id-1][1]=t;
        }
        memset(mark,0,sizeof(mark));
        creat(1,1,t);
        for(i=0;i<n;i++)
        {
            int l=pos[i][0];
            int r=pos[i][1];
            update(1,l,r,i);
        }
        ans=0;
        query(1,1,t);
        printf("%d\n",ans);
    }
    return 0;
}

poj 2528 Mayor's posters （线段树+区间离散）