题目大意:在数轴上有$n$个闭区间$[l_1,r_1],[l_2,r_2],...,[l_n,r_n]$。现在要从中选出 $m$ 个区间,使得这 $m$ 个区间共同包含至少一个位置。输出被选中的最长区间长度减去被选中的最短区间长度,若多解,输出最小的一个
题解:把区间按长度排序,然后把左右端点离散化,双指针扫一下,线段树维护一下有没有点被覆盖$ \geq m$次即可
卡点:1.因为有$2\times n$个点,所以线段树开小了一倍
C++ Code:
#include <cstdio>
#include <algorithm>
#define maxn 500010
using namespace std;
const int inf = 0x3f3f3f3f;
int n, m, tot, M, ans;
int p[maxn << 1];
struct interval {
int l, r, len;
bool operator < (const interval & b) const {return len < b.len;}
} s[maxn];
inline int min(int a, int b) {return a < b ? a : b;}
inline int max(int a, int b) {return a > b ? a : b;}
int V[maxn << 3], cov[maxn << 3];
void pushdown(int rt) {
int &tmp = cov[rt];
V[rt << 1] += tmp;
V[rt << 1 | 1] += tmp;
cov[rt << 1] += tmp;
cov[rt << 1 | 1] += tmp;
tmp = 0;
}
void add(int rt, int l, int r, int L, int R, int num = 1) {
if (L <= l && R >= r) {
V[rt] += num;
cov[rt] += num;
return ;
}
if (cov[rt]) pushdown(rt);
int mid = l + r >> 1;
if (L <= mid) add(rt << 1, l, mid, L, R, num);
if (R > mid) add(rt << 1 | 1, mid + 1, r, L, R, num);
V[rt] = max(V[rt << 1], V[rt << 1 | 1]);
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%d%d", &s[i].l, &s[i].r);
s[i].len = s[i].r - s[i].l;
p[i - 1 << 1 | 1] = s[i].l;
p[i << 1] = s[i].r;
}
tot = n << 1;
sort(p + 1, p + tot + 1);
sort(s + 1, s + n + 1);
M = lower_bound(p + 1, p + tot + 1, p[tot]) - p;
for (int i = 1; i <= n; i++) {
s[i].l = lower_bound(p + 1, p + tot + 1, s[i].l) - p;
s[i].r = lower_bound(p + 1, p + tot + 1, s[i].r) - p;
}
int L = 1;
ans = inf;
for (int i = 1; i <= n; i++) {
add(1, 1, M, s[i].l, s[i].r);
while (V[1] >= m && L <= i) {
ans = min(ans, s[i].len - s[L].len);
add(1, 1, M, s[L].l, s[L].r, -1);
L++;
}
}
printf("%d\n", (ans == inf) ? -1 : ans);
return 0;
}