标签：nested lin get empty \n turn pre www. sts
http://codeforces.com/contest/1029/problem/C
You are given 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 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.
The first line contains a single integer nn (2≤n≤3?1052≤n≤3?105) — the number of segments in the sequence.
Each of the next nn lines contains two integers lili and riri (0≤li≤ri≤1090≤li≤ri≤109) — the description of the ii-th segment.
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.
4
1 3
2 6
0 4
3 3
1
5
2 6
1 3
0 4
1 20
0 4
2
3
4 5
1 2
9 20
0
2
3 10
1 5
7
代码：
#include <bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
const int maxn = 300010 + 10;
int N;
struct Node {
int l;
int r;
}S[maxn], Q[maxn], A[maxn];
int main() {
scanf("%d", &N);
S[0].r = inf, S[0].l = -inf;
for(int i = 1; i <= N; i ++) {
scanf("%d%d", &A[i].l, &A[i].r);
S[i].l = max(S[i - 1].l, A[i].l);
S[i].r = min(S[i - 1].r, A[i].r);
}
Q[N + 1].r = inf, Q[N + 1].l = -inf;
for(int i = N; i >= 1; i --) {
Q[i].l = max(A[i].l, Q[i + 1].l);
Q[i].r = min(A[i].r, Q[i + 1].r);
}
int ans = 0;
for(int i = 1; i <= N; i ++) {
ans = max(ans, (min(Q[i + 1].r, S[i - 1].r) - max(Q[i + 1].l, S[i - 1].l)));
}
printf("%d\n", ans);
return 0;
}
CodeForces C. Maximal Intersection
标签：nested lin get empty \n turn pre www. sts
原文地址：https://www.cnblogs.com/zlrrrr/p/9822794.html
