poj 2528 （线段树+离散化）

poj 2528

 

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

 

题意：贴报纸，可以互相覆盖，求最后能看见多少。

数据很大，不离散会超出内存。将浪费的部分去掉，将出现过的所有点其映射到相距更近的点上。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define N 100005
#define mod 258280327
#define MIN 0
#define MAX 1000001

int l[N],r[N];
int x[N];
int has[10000005];

struct node
{
    int le,re;
    bool covered;
} pnode[10*N];

void build(int i,int l,int r)
{
    pnode[i].le = l;
    pnode[i].re = r;
    pnode[i].covered = false;
    if(l == r)
        return;

    build(2 * i,l,(l+r)/2);
    build(2*i+1,(l+r)/2+1,r);
}

bool insert(int i,int l,int r,int a,int b)//是否可见
{
    if(pnode[i].covered == true)//如果要找的已经被覆盖
        return false;
    if(l == a && r == b)
    {
        pnode[i].covered = true;
        return true;
    }

    bool ans;
    int mid = (l + r)>>1;
    if (mid >= b)
        ans = insert(i*2,l, mid,  a, b);
    else if (mid < a)
        ans = insert(i*2+1,mid+1, r, a, b);
    else
    {
        bool x1 = insert(i*2,l, mid, a, mid);
        bool x2 = insert(i*2+1,mid+1, r, mid+1, b);
        ans = x1 || x2;
    }

    if(pnode[i*2].covered == true && pnode[i*2+1].covered == true)
        pnode[i].covered = true;
    return ans;
}




int main()
{
    int T,n,tot;
    scanf("%d",&T);

    while(T--)
    {
        scanf("%d",&n);
        tot = 0;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d%d",&l[i],&r[i]);
            x[tot++] = l[i];
            x[tot++] = r[i];
        }

        sort(x,x+tot);
        tot = unique(x,x+tot)-x;
        int all=1;
        for(int i = 0;i < tot;i++)
        {
            has[x[i]] = all;
            if(i < tot - 1)
            {
                if(x[i+1] - x[i] == 1)
                    all++;
                else
                    all+=2;
            }
        }

        build(1,1,all);
        int num = 0;

        for(int i = n;i >=1;i--)
        {
            if(insert(1,1,all,has[l[i]],has[r[i]]))
                num ++;
        }

        printf("%d\n",num);

    }
    return 0;
}