POJ-2528-Mayor's posters

链接:https://vjudge.net/problem/POJ-2528#author=swust20141567

题意:

n(n<=10000) 个人依次贴海报,给出每张海报所贴的范围li,ri(1<=li<=ri<=10000000) 。求出最后还能看见多少张海报。

思路:

离散化加线段树。

离散化降低区间,并且因为区间重叠,每两个差距大于1的位置中间加一个前一个数+1的数,确保答案正确。

代码:

#include <iostream>
#include <memory.h>
#include <vector>
#include <map>
#include <algorithm>

using namespace std;

typedef long long LL;

const int MAXN = 1e5 + 10;
struct Line
{
    int _l, _r;
}line[MAXN];
int num[MAXN * 4];
int segment[MAXN * 4];
int vis[MAXN];

void Push_down(int root)
{
    if (segment[root])
    {
        segment[root << 1] = segment[root];
        segment[root << 1 | 1] = segment[root];
        segment[root] = 0;
    }
}

void Update_tree(int root, int l, int r, int ql, int qr, int c)
{
    if (ql > r || qr < l)
        return ;
    if (ql <= l && r <= qr)
    {
        segment[root] = c;
        return ;
    }
    int mid = (l + r) / 2;
    Push_down(root);
    Update_tree(root << 1, l, mid, ql, qr, c);
    Update_tree(root << 1 | 1, mid + 1, r, ql, qr, c);
}

int Query(int root, int l, int r, int w)
{
    if (l == r)
        return segment[root];
    int mid = (l + r) / 2;
    Push_down(root);
    if (w <= mid)
        return Query(root << 1, l, mid, w);
    else
        return Query(root << 1 | 1,mid + 1, r, w);
}

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        memset(vis, 0, sizeof(vis));
        memset(num, 0, sizeof(num));
        memset(segment, 0, sizeof(segment));
        int n;
        scanf("%d", &n);
        int cnt = 0;
        for (int i = 1;i <= n;i++)
        {
            scanf("%d%d", &line[i]._l, &line[i]._r);
            num[++cnt] = line[i]._l;
            num[++cnt] = line[i]._r;
        }
        sort(num + 1, num + 1 + cnt);
        int size = unique(num + 1, num + 1 + cnt) - (num + 1);//可用数目
        int sum = size;
        for (int i = 2;i <= size;i++)
            if (num[i] - num[i - 1] > 1)
                num[++sum] = num[i - 1] + 1;
        sort(num + 1, num + 1 + sum);//处理过后的数组
        for (int i = 1;i <= n;i++)
        {
            int ll = lower_bound(num + 1, num + 1 + sum, line[i]._l) - num;
            int rr = lower_bound(num + 1, num + 1 + sum, line[i]._r) - num;
            Update_tree(1, 1, sum, ll, rr, i);
        }
        int res = 0;
        for (int i = 1;i <= sum;i++)
            vis[Query(1, 1, sum, i)] = 1;
        for (int i = 1;i <= sum;i++)
            res += vis[i];
        printf("%d\n", res);
    }

    return 0;
}

  

posted @ 2019-02-24 15:35  YDDDD  阅读(168)  评论(0编辑  收藏  举报