E. Sign on Fence time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Bizon the Champion has recently finished painting his wood fence. The fence consists of a sequence of n panels of 1 meter w…

Bizon the Champion has recently finished painting his wood fence. The fence consists of a sequence of n panels of 1 meter width and of arbitrary height. The i-th panel's height is hi meters. The adjacent planks follow without a gap between them. Afte…

题目链接:Codeforces 484E Sign on Fence 题目大意:给定给一个序列,每一个位置有一个值,表示高度,如今有若干查询,每次查询l,r,w,表示在区间l,r中, 连续最长长度大于w的最大高度为多少. 解题思路:可持久化线段树维护区间合并,前端时间碰到一题可持久化字典树,就去查了一下相关论文,大概知道了是 什么东西. 将高度依照从大到小的顺序排序,然后每次插入一个位置,线段树维护最长连续区间,由于插入是依照从大到小的顺 序,所以每次的线段树中的连续最大长度都是满足高度大于等于…

http://codeforces.com/contest/484/problem/E 题意: 给出n个数,查询最大的在区间[l,r]内,长为w的子区间的最小值 第i棵线段树表示>=i的数 维护最长连续子区间 把数从大到小插入主席树 对于每个询问,二分x 在第x棵线段树中查,若最长连续子区间>=w,到代表更大的线段树中查 没有建第n+1棵线段树,导致前面节点的siz不对,WA了一次 #include<cstdio> #include<iostream> #include…

E. Sign on Fence time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Bizon the Champion has recently finished painting his wood fence. The fence consists of a sequence of n panels of 1 meter w…

[CF484E]Sign on Fence(主席树) 题面 懒得贴CF了,你们自己都找得到 洛谷 题解 这不就是[TJOI&HEOI 排序]那题的套路吗... 二分一个答案,把大于答案的都变成\(1\),其余变成\(0\) 按照题目要求的区间内连续的\(K\)个 就是检查最长的连续\(1\)的子段长度大于\(K\) 所以维护\(1\)的子段长度(这也是原题吧??) 因为范围比较大,不能每次开线段树计算 我们发现每次将范围增大的时候,在线段树上可以直接做一定的修改 又因为要维护所有的线段树,所以直…

E. Sign on Fence   Bizon the Champion has recently finished painting his wood fence. The fence consists of a sequence of n panels of 1 meter width and of arbitrary height. The i-th panel's height is hi meters. The adjacent planks follow without a gap…

CF484E Sign on Fence #include<bits/stdc++.h> #define RG register #define IL inline #define _ 100100 #define inf 1e9+7 using namespace std; IL int gi(){ RG int data = 0 , m = 1; RG char ch = 0; while(ch != '-' && (ch<'0' || ch > '9')) c…

http://codeforces.com/problemset/problem/484/E 题意: 给定一个长度为n的数列,有m次询问,询问形如l r k 要你在区间[l,r]内选一个长度为k的区间,求区间最小数的最大值 整体二分啊....O((n+m)log(n)log(值域)) 对于一个询问,可以二分答案后(设二分出的答案为ans),将>=ans的数标记为1,<ans的数标记为0,然后求该询问区间内最长的连续且全为1的一段的长度x,如果x>=该询问要求的k则合法,否则不合法: 多次…

http://codeforces.com/contest/349/problem/B 贪心 #include <cstdio> #include <cstring> #include <algorithm> using namespace std; <<; ]; int v,k; int main() { while(scanf("%d",&v)!=EOF) { int min1=inf; k=; ; i<=; i++)…