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步骤
1.先将所有区间按左端点排序
2.从前往后扫一遍，维护当前正在合并的区间[st, ed]
3.依次检查每个区间[l, r]，
若 l > ed,将[st, ed]加入 ans , 更新st = l,ed = r
若 l <= ed ，更新ed = max(ed, r)
时间复杂度 O(nlogn)
模板
https://www.acwing.com/problem/content/description/424/
code：
l, m = map(int, input().split())
grid = [list(map(int, input().split())) for _ in range(m)]
grid.sort()
p = grid[0]
st, ed = p[0], p[1]
sum = 0
for p in grid[1:]: if ed < p[0]: sum += ed - st + 1 st, ed = p[0], p[1] else: ed = max(ed, p[1])
sum += ed - st + 1
print(l + 1 - sum)
小试牛刀
https://www.acwing.com/problem/content/1345/
code：
n = int(input())
grid = [list(map(int, input().split())) for _ in range(n)]
grid.sort()
p = grid[0]
st, ed = p[0], p[1]
ans, res = 0, 0
for p in grid[1:]: if ed < p[0]: ans = max(ans, ed - st) res = max(res, p[0] - ed) st, ed = p[0], p[1] else: ed = max(ed, p[1])
ans = max(ans, ed - st)
print(ans, res)
真题演练
第十四届蓝桥杯省赛Python B 组 D 题
二分答案 + 区间合并
AC_code:
import sys
input = sys.stdin.readline
inf = float('inf') n, length = map(int, input().split())
grid = [list(map(int, input().split())) for _ in range(n)] def check(mid): a = [] for pos, t in grid: if t > mid: continue l = max(1, pos - (mid - t)) r = min(length, pos + (mid - t)) a.append((l, r)) a.sort() if not a: return 0 st, ed = a[0][0], a[0][1] if st != 1: return 0 for p in a[1:]: if ed + 1 < p[0]: return 0 else: ed = max(ed, p[1]) if ed != length: return 0 return 1 l, r = 1, 10**18
while l <= r: mid = (l + r) // 2 if check(mid): ans = mid r = mid - 1 else: l = mid + 1 print(ans)
END
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