题目链接:传送门
lr 1e9
离散化是必须的
离散化之前把区间长度记下来就行
而且lr坐标要一起离散化
只要被覆盖的区间>=m就不断删除区间取更小答案
/**
* @Date: 2019-03-31T15:18:21+08:00
* @Last modified time: 2019-03-31T15:18:22+08:00
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <complex>
#include <algorithm>
#include <climits>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define
#define
using namespace std;
typedef long long ll;
struct node {
int l, r, w, f;
}tree[A];
struct Node {
int l, r, len;
}e[A];
int n, m, a, b, c, t[A], tn;
void build(int k, int l, int r) {
tree[k].l = l; tree[k].r = r;
if (l == r) {
tree[k].w = tree[k].f = 0;
return;
}
int m = (l + r) >> 1;
build(k << 1, l, m);
build(k << 1 | 1, m + 1, r);
}
void down(int k) {
tree[k << 1].f += tree[k].f;
tree[k << 1 | 1].f += tree[k].f;
tree[k << 1].w += tree[k].f;
tree[k << 1 | 1].w += tree[k].f;
tree[k].f = 0;
}
void change(int k, int l, int r, int add) {
if (tree[k].l >= l and tree[k].r <= r) {
tree[k].w += add;
tree[k].f += add;
return;
}
if (tree[k].f) down(k);
int m = (tree[k].l + tree[k].r) >> 1;
if (l <= m) change(k << 1, l, r, add);
if (r > m) change(k << 1 | 1, l, r, add);
tree[k].w = max(tree[k << 1].w, tree[k << 1 | 1].w);
}
int main(int argc, char const *argv[]) {
cin >> n >> m;
for (int i = 1; i <= n; i++) {
cin >> e[i].l >> e[i].r;
e[i].len = e[i].r - e[i].l;
t[++tn] = e[i].l, t[++tn] = e[i].r;
}
sort(t + 1, t + tn + 1);
tn = unique(t + 1, t + tn + 1) - t - 1;
for (int i = 1; i <= n; i++) {
e[i].l = lower_bound(t + 1, t + tn + 1, e[i].l) - t;
e[i].r = lower_bound(t + 1, t + tn + 1, e[i].r) - t;
}
sort(e + 1, e + n + 1, [] (Node a, Node b) -> bool {return a.len < b.len;});
int ans = 0x3f3f3f3f; build(1, 1, tn);
for (int i = 1, j = 1; i <= n; i++) {
change(1, e[i].l, e[i].r, 1);
while (tree[1].w >= m) {
change(1, e[j].l, e[j].r, -1);
ans = min(ans, e[i].len - e[j].len);
j++;
}
}
if (ans == 0x3f3f3f3f) puts("-1");
else cout << ans << endl;
}