poj 2528(区间修改+离散化)

题意：有一个黑板上贴海报，给出每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：因为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，比如[1,10]，[1,4]，[6,10]普通得到答案是2，但其实是3，改进的离散化方法如果两个数字相差大于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
, r
, a[N << 3], tree[N << 4], vis
, 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;
}