zoj3686(线段树的区间更新)】的更多相关文章

对线段树的区间更新有了初步的了解... A Simple Tree Problem Time Limit: 3 Seconds      Memory Limit: 65536 KB Given a rooted tree, each node has a boolean (0 or 1) labeled on it. Initially, all the labels are 0. We define this kind of operation: given a subtree, negat…
Color the ball Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7941    Accepted Submission(s): 4070 Problem Description N个气球排成一排,从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球…
Just a Hook Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15129    Accepted Submission(s): 7506 Problem Description In the game of DotA, Pudge’s meat hook is actually the most horrible thing f…
题意: long long data[250001]; void A( int st, int nd ) { for( int i = st; i \le nd; i++ ) data[i] = data[i] + (i - st + 1); } void B( int st, int nd ) { for( int i = st; i \le nd; i++ ) data[i] = data[i] + (nd - i + 1); } void C( int st, int nd, int x…
Just a Hook Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 13001    Accepted Submission(s): 6476 Problem Description In the game of DotA, Pudge’s meat hook is actually the most horrible thing…
http://acm.hdu.edu.cn/showproblem.php?pid=1556 题意: N个气球,每次[a,b]之间的气球涂一次色,统计每个气球涂色的次数. 思路: 这道题目用树状数组和线段树都可以,拿这道题来入门一下线段树的区间更新. #include<iostream> #include<cstring> #include<algorithm> using namespace std; + ; int n; int ans[maxn]; struct…
N个气球排成一排,从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球a开始到气球b依次给每个气球涂一次颜色.但是N次以后lele已经忘记了第I个气球已经涂过几次颜色了,你能帮他算出每个气球被涂过几次颜色吗? Input每个测试实例第一行为一个整数N,(N <= 100000).接下来的N行,每行包括2个整数a b(1 <= a <= b <= N). 当N = 0,输入结束.Output每个测…
Color the Ball Time Limit: 2 Seconds      Memory Limit: 65536 KB There are infinite balls in a line (numbered 1 2 3 ....), and initially all of them are paint black. Now Jim use a brush paint the balls, every time give two integers a b and follow by…
绝对是很好的题 把问题转化成当第i个询问的答案是数值x时是否可行 要判断值x是否可行,只要再将问题转化成a数组里>=x的值数量是否严格大于b数组里的>=x的值 那么线段树叶子结点维护对于值x的a数组里的合法数数量-b数组里的合法数数量,如果是正数即这个值可行 线段树维护区间最大值,然后询问最靠右的非负叶子下标 #include<bits/stdc++.h> #include<vector> using namespace std; #define maxn 100000…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3397 题意:给你一个长度为n的0,1序列,支持下列五种操作, 操作0(0 a b):将a到b这个区间的元素全部置为0. 操作1(1 a b):将a到b这个区间的元素全部置为1. 操作2(2 a b):将a到b这个区间所有的0置为1,所有的1置为0. 操作3(3 a b):查询a到b这个区间1的总数. 操作4(4 a b):查询a到b这个区间连续1的最长长度 本题属于简单的区间更新线段树 重点:0操作…