CF1029C Maximal Intersection 暴力枚举

Maximal Intersection

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given $\mathrm{nn segments\; on\; a\; number\; line;\; each\; endpoint\; of\; every\; segment\; has\; integer\; coordinates.\; Some\; segments\; can\; degenerate\; to\; points.\; Segments\; can\; intersect\; with\; each\; other,\; be\; nested\; in\; each\; other\; or\; even\; coincide.}$

The intersection of a sequence of segments is such a maximal set of points (not necesserily having integer coordinates) that each point lies within every segment from the sequence. If the resulting set isn't empty, then it always forms some continuous segment. The length of the intersection is the length of the resulting segment or $\mathrm{00 in\; case\; the\; intersection\; is\; an\; empty\; set.}$

For example, the intersection of segments $[1;5][1;5] and [3;10][3;10] is [3;5][3;5] (length 22),\; the\; intersection\; of\; segments [1;5][1;5] and [5;7][5;7] is [5;5][5;5] (length 00)\; and\; the\; intersection\; of\; segments [1;5][1;5] and [6;6][6;6] is\; an\; empty\; set\; (length 00).$

Your task is to remove exactly one segment from the given sequence in such a way that the intersection of the remaining $(n-1)(n-1) segments\; has\; the\; maximal\; possible\; length.$

Input

The first line contains a single integer $\mathrm{nn (2\le n\le 3\cdot 1052\le n\le 3\cdot 105)\; —\; the\; number\; of\; segments\; in\; the\; sequence.}$

Each of the next $\mathrm{nn lines\; contains\; two\; integers lili and riri (0\le li\le ri\le 1090\le li\le ri\le 109)\; —\; the\; description\; of\; the ii-th\; segment.}$

Output

Print a single integer — the maximal possible length of the intersection of $(n-1)(n-1) remaining\; segments\; after\; you\; remove\; exactly\; one\; segment\; from\; the\; sequence.$

Examples

input

Copy

4
1 3
2 6
0 4
3 3

output

Copy

1

input

Copy

5
2 6
1 3
0 4
1 20
0 4

output

Copy

2

input

Copy

3
4 5
1 2
9 20

output

Copy

0

input

Copy

2
3 10
1 5

output

Copy

7

Note

In the first example you should remove the segment $[3;3][3;3],\; the\; intersection\; will\; become [2;3][2;3] (length 11).\; Removing\; any\; other\; segment\; will\; result\; in\; the\; intersection [3;3][3;3] (length 00).$

In the second example you should remove the segment $[1;3][1;3] or\; segment [2;6][2;6],\; the\; intersection\; will\; become [2;4][2;4] (length 22)\; or [1;3][1;3] (length 22),\; respectively.\; Removing\; any\; other\; segment\; will\; result\; in\; the\; intersection [2;3][2;3] (length 11).$

In the third example the intersection will become an empty set no matter the segment you remove.

In the fourth example you will get the intersection $[3;10][3;10] (length 77)\; if\; you\; remove\; the\; segment [1;5][1;5] or\; the\; intersection [1;5][1;5] (length 44)\; if\; you\; remove\; the\; segment [3;10][3;10].$

题意：已知n个区间，求删去一个区间后剩余的区间交的区域最大值

分析：求去掉某段区间后剩余的区间的相交区域，相当于求这个区间前面所有区间的相交区域和这个区间后的所有相交区域的交集

　　在开始的时候枚举出所有点，关于这点前的相交区域pre[i-1]，关于这点后的相交区域pos[i+1]

　　然后枚举pre[i-1]与pos[i+1]相交区域的最大值

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 3e5+10;
const double eps = 1e-8;
const ll mod = 1e9 + 7;
const ll inf = 1e9;
const double pi = acos(-1.0);
struct node {
    ll x, y;
};
node a[maxn], pre[maxn], pos[maxn];
int main() {
    ll n, maxle1 = 0, minri1 = 1e9+10, maxle2 = 0, minri2 = 1e9+10;
    scanf("%lld",&n);
    pos[n+1].y = pre[0].y = 1e9+10;
    for( ll i = 1; i <= n; i ++ ) {
        scanf("%lld%lld",&a[i].x,&a[i].y);
        maxle1 = max(maxle1,a[i].x), minri1 = min(minri1,a[i].y);
        pre[i].x = maxle1, pre[i].y = minri1;
    }
    for( ll i = n; i >= 1; i -- ) {
        maxle2 = max(maxle2,a[i].x), minri2 = min(minri2,a[i].y);
        pos[i].x = maxle2, pos[i].y = minri2;
    }
    ll ans = 0;
    for( ll i = 1; i <= n; i ++ ) {
        ll le = max(pre[i-1].x,pos[i+1].x), ri = min(pre[i-1].y,pos[i+1].y);
        ans = max(ans,ri-le);
    }
    printf("%lld
",ans);
    return 0;
}