随笔分类 - 线段树or树状数组
摘要:Just a Hook Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 25126 Accepted Submission(s): 12545
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摘要:Coder Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4838 Accepted Submission(s): 1853 Proble
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摘要:Man Down Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2030 Accepted Submission(s): 743 Proble
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摘要:Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 16046 Accepted Submissi
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摘要:I Hate It Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 57279 Accepted Submission(s): 22365 Pr
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摘要:敌兵布阵 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 66634 Accepted Submission(s): 28074 Problem
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摘要:Rikka with Phi Accepts: 5 Submissions: 66 Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Rikka with Phi Accepts:
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摘要:Dynamic Rankings Time Limit: 10000MS Memory Limit: 32768KB 64bit IO Format: %lld & %llu Submit Status Description The Company Dynamic Rankings has dev
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摘要:D-query Time Limit: 227MS Memory Limit: 1572864KB 64bit IO Format: %lld & %llu Submit Status Description English Vietnamese Given a sequence of n numb
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摘要:题意: 输入描述 输出描述 输入样例 输出样例 思路: a[i]表示[1,i]的逆序对,所以a[i] - a[i-1]便是i前面比第i个数大的,所以a[i]-a[i-1]+1便是第i个数在[1,i]中第几大。 所以用树桩数组全部初始化为1,然后二分+树状数组找出第k大的数然后把其赋值为0即可。
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摘要:题意:在一个队列中,你知道一个人在他左边或者右边比他高的人的个数,求字典序最小的答案 思路:先将人按 矮-->高 排序,然后算出在每个人前面需要预留的位置。树状数组(也可以线段树)解决时,先二分查找合适的位置,如果合理则标记,每次查找时 枚举的人数 - 前面已占的人数 = 可预留位置 判断分别在它左
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摘要:题意:基于矩阵的NIM游戏,求异或和。 思路:在x1,y1 到 x2, y2的异或和 = A[ x2 ][ y2 ] ^ A[x1-1][ y2 ] ^ A[ x2 ][y1 - 1] ^ A[ x1-1 ][ y1 - 1 ] 先普通来了两次,结果都超时。 上个二维树状数组AC
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摘要:题意: 两个操作:① 当为1时 ,乘上后面的数 ② 当为2时,除以第x次乘的数 还说了2操作后面的n不会重复(就这明显看出线段树- -,然而并没有看出来,还是靠的队友) 1则对每个节点赋值,2则将相应的点改为一,每次操作后进行一次询问
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摘要:poj 2528 For each input data set print the number of visible posters after all the posters are placed. The picture below illustrates the case of the s
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摘要:poj 3264 Sample Input Sample Output 求任一区间的最大值和最小值的差 poj 3468 Sample Input Sample Output ①对区间i - j 的数全加上c ; ②求区间的和
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摘要:Sample Input 3 0 0 0 3 0 1 5 0 1 0 0 1 1 0 1 0 0 3 0 0 0 3 0 1 5 0 1 0 0 1 1 0 1 0 0 Sample Output Case #1: 0 0 0 Case #2: 0 1 0 2 Case #1: 0 0 0 Case
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摘要:Fantasy magicians usually gain their ability through one of three usual methods: possessing it as an innate talent, gaining it through study and pract
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摘要:转自:http://blog.csdn.net/x314542916/article/details/7837276(学习下) 线段树的入门级 总结 线段树是一种二叉搜索树,与区间树相似,它将一个区间划分成一些单元区间,每个单元区间对应线段树中的一个叶结点。 对于线段树中的每一个非叶子节点[a,b]
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