小小粉丝度度熊 二分答案 + two pointer
http://acm.hdu.edu.cn/showproblem.php?pid=6119
发现自己的two pointer能力超弱。
这题是合并时间后,二分答案。
可以知道对于每个时间区间,合法的答案肯定是从其开始时间,向左扩展
或者从其结束时间,向右扩展。
复杂度O(2n log val)
细节有点恶心
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; const int maxn = 100000 + 20; struct Node { int be, en; bool operator < (const struct Node & rhs) const { if (be != rhs.be) return be < rhs.be; else return en < rhs.en; } }node[maxn]; int m, n; LL sum[maxn]; bool check(LL val) { int p1 = 1, p2 = 1; while (p1 <= n) { while (p2 <= n && node[p2].en - node[p1].be + 1 < val) { p2++; } if (p2 == n + 1) { LL cut = sum[n] - sum[p1]; cut += node[p1].be + val - 1 - node[n].en; if (cut <= m) return true; p1++; continue; } // if (node[p2].en - node[p1].be + 1 < val) break; if (node[p2].be - node[p1].be + 1 > val) { if (sum[p2 - 1] - sum[p1] + node[p1].be + val - 1 - node[p2 - 1].en <= m) return true; } else { if (sum[p2] - sum[p1] <= m) return true; } p1++; } p1 = p2 = 1; while (p2 <= n && node[p2].en < val) p2++; while (p2 <= n) { while (p1 <= n && node[p2].en - node[p1].en + 1 > val) { p1++; } if (node[p2].en - node[p1].be + 1 >= val) { if (sum[p2] - sum[p1] <= m) return true; } else { LL cut = sum[p2] - sum[p1]; LL to = node[p2].en - val + 1; cut += node[p1].be - to; if (cut <= m) return true; } p2++; } return false; } void work() { for (int i = 1; i <= n; ++i) { scanf("%d%d", &node[i].be, &node[i].en); } sort(node + 1, node + 1 + n); int len = 1; for (int i = 2; i <= n; ++i) { if (node[len].en >= node[i].be) node[len].en = max(node[len].en, node[i].en); else { node[++len] = node[i]; } } n = len; for (int i = 2; i <= n; ++i) { sum[i] = sum[i - 1] + node[i].be - node[i - 1].en - 1; } LL be = 0, en = 1e15; for (int i = 1; i <= n; ++i) { be = max(be, 1LL * node[i].en - node[i].be + 1); } while (be <= en) { LL mid = (be + en) >> 1; if (check(mid)) be = mid + 1; else en = mid - 1; } cout << en << endl; } int main() { #ifdef local freopen("data.txt", "r", stdin); // freopen("data.txt", "w", stdout); #endif while (scanf("%d%d", &n, &m) > 0) work(); return 0; }
既然选择了远方,就要风雨兼程~
posted on 2017-08-14 21:53 stupid_one 阅读(284) 评论(0) 编辑 收藏 举报