CodeForces C. Maximal Intersection

最新推荐文章于 2024-04-26 10:48:25 发布

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最新推荐文章于 2024-04-26 10:48:25 发布

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原文链接：http://www.cnblogs.com/zlrrrr/p/9822794.html

本文解析了Codeforces竞赛中一道关于线段交集长度最大化的题目，通过构造辅助数组和遍历策略，实现了高效求解算法。示例输入输出展示了算法的有效性。

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 $(n-1)(n-1)segments has the maximal possible length.$

Input

The first line contains a single integer $nn (2\leqn\leq3\cdot1052\leqn\leq3\cdot105) — the number of segments in the sequence.$

Each of the next $nn lines contains two integers lili and riri (0\leqli\leqri\leq1090\leqli\leqri\leq109) — the description of the ii-th segment.$

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

output

Copy

input

Copy

output

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input

Copy

output

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input

Copy

2
3 10
1 5

output

Copy

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;
}

转载于:https://www.cnblogs.com/zlrrrr/p/9822794.html