poj 2528(区间改动+离散化)

Posted 2020-09-29 cxchanpin

题意：有一个黑板上贴海报。给出每一个海报在黑板上的覆盖区间为l r，问最后多少个海报是可见的。

题解：由于l r取值到1e7，肯定是要离散化的，但普通的离散化会出问题。比方[1,10]，[1,4]，[6,10]普通得到答案是2，但事实上是3。改进的离散化方法假设两个数字相差大于1，就在中间补一个数字。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 10005;
int n, l[N], r[N], a[N << 3], tree[N << 4], vis[N], res;

void pushdown(int k) {
    if (tree[k] != -1) {
        tree[k * 2] = tree[k * 2 + 1] = tree[k];
        tree[k] = -1;
    }
}

void modify(int k, int left, int right, int l1, int r1, int x) {
     if (l1 <= left && right <= r1) {
         tree[k] = x;
         return;
     }
     pushdown(k);
     int mid = (left + right) / 2;
     if (mid >= l1)
        modify(k * 2, left, mid, l1, r1, x);
    if (mid < r1)
        modify(k * 2 + 1, mid + 1, right, l1, r1, x);
}

void query(int k, int left, int right) {
    if (left == right) {
        if (!vis[tree[k]]) {
            res++;
            vis[tree[k]] = 1;
       }
       return;
    }
    pushdown(k);
    int mid = (left + right) / 2;
    query(k * 2, left, mid);
    query(k * 2 + 1, mid + 1, right);
}

int main() {
    int t;
    scanf ("%d", &t);
    while (t--) {
        memset(tree, -1, sizeof(tree));
        memset(vis, 0, sizeof(vis));
        int cnt = 0;
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
             scanf ("%d%d", &l[i], &r[i]);
             a[++cnt] = l[i];
             a[++cnt] = r[i];
        }
        sort(a + 1, a + 1 + cnt);
        cnt = unique(a + 1, a + 1 + cnt) - (a + 1);
        int cnt2 = cnt;
        for (int i = 2; i <= cnt; i++)
            if (a[i] - a[i - 1] > 1)
                a[++cnt2] = a[i] - 1;
        cnt = cnt2;
        sort(a + 1, a + 1 + cnt);
        for (int i = 1; i <= n; i++) {
            int l1 = lower_bound(a + 1, a + 1 + cnt, l[i]) - a;
            int r1 = lower_bound(a + 1, a + 1 + cnt, r[i]) - a;
            modify(1, 1, cnt, l1, r1, i);
        }
    res = 0;
    query(1, 1, cnt);
        printf("%d\n", res);
    }
    return 0;
}