链接: 对于POJ老是爆,我也是醉了, 链接等等再发吧! http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82832#problem/G 只是简单的建树,每个节点上记录它的最大值和最小值,最后查询一下,就ok了 代码: #include<cstdio> #include<cmath> #include<cstring> #include<cstdlib> #include<algorithm&…

For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows fr…

题意 给出一个序列  每次查询区间的max-min是多少 思路:直接维护max 和min即可  写两个query分别查最大最小值 #include<cstdio> #include<algorithm> #include<set> #include<vector> #include<cstring> #include<iostream> using namespace std; ; int a[maxn]; struct Node{…

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define MAXN 50010 int n , q; int h[MAXN]; struct Node{ int l; int r; int minNum; int maxNum; } tr[*MAXN]; void build(int l,int r,int u)…

这题,看到别人的解题报告做出来的,分析: 大概意思就是: 数组sum[i][j]表示从第1到第i头cow属性j的出现次数. 所以题目要求等价为: 求满足 sum[i][0]-sum[j][0]=sum[i][1]-sum[j][1]=.....=sum[i][k-1]-sum[j][k-1] (j<i) 中最大的i-j 将上式变换可得到 sum[i][1]-sum[i][0] = sum[j][1]-sum[j][0] sum[i][2]-sum[i][0] = sum[j][2]-sum[j]…

Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For example, cows exhibiting feature #1…

/******************************************************* 题目: Balanced Lineup(poj 3264) 链接: http://poj.org/problem?id=3264 题意: 给个数列,查询一段区间的最大值与最小值的差 算法: RMQ ********************************************************/ #include<cstdio> #include<cstrin…

G - Balanced Lineup POJ - 3264 思路:水题,线段树的基本操作即可. #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 50010 using namespace std; int n,q; struct nond{ int l,r,min,max; }tree[MAXN*]; void up(int now…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 53703   Accepted: 25237 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 68466   Accepted: 31752 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh…