题面: http://poj.org/problem?id=2329 题解: 这题有很多做法 1. 搜索 复杂度$O(n^4)$ 但是实际上远远达不到这个复杂度 所以可以通过 2. 对于每一个格子,我们枚举每一行,找到每一行离他最近的格子 当前格子向右移动时,每一行离他最近的格子不可能向左移动,所以复杂度$O(n^3)$ 3. dp 一共四个方向 左上到右下 左下到右上 右上到左下 右下到左上 然后分别dp 找到这个方向离他最近的格子 以左上到右下为例 $f[i][j]$可由$f[i-1][j]…

Link: POJ 2329 传送门 Solution: 比较明显的$dp$,但爆搜好像也能过 用多个方向$dp$来解决此题,最后汇总答案即可 一开始我写了4个,但后来发现只要相反的2个方向即可,同时不用分别记录答案,直接不断更新答案即可 要特别注意对特例的判断: 不能只判断其最近距离相同且最近点相同! 仅当$(a1,b1)$和$(a2,b2)$当前都仅有一个最近点且其相同时才不增加权值 否则可能$(a2,b2)$有多个最近点但正好记录了与$(a1,b1)$最近点相同的点 Code: #incl…

题目 传送门:QWQ 分析 在dp分类里做的,然而并不会$ O(n^3) $ 的$ dp $,怒写一发搜索. 看起来是$ O(n^4) $,但仔细分析了一下好像还挺靠谱的? poj挂了,没在poj交,在zoj上交的500ms P.S. 如果要在poj交还要把多数据改成单数据 代码 #include <bits/stdc++.h> using namespace std; ; ], cnt, ok[maxn][maxn], n; int now[maxn][maxn], ans[maxn][ma…

POJ - 1330 Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000KB   64bit IO Format: %lld & %llu Submit Status Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:  In t…

POJ 1330 Nearest Common Ancestors / UVALive 2525 Nearest Common Ancestors (最近公共祖先LCA) Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each node is labeled with an…

POJ 1330 Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24209   Accepted: 12604 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:  In the fi…

POJ 1330 Nearest Common Ancestors A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:  In the figure, each node is labeled with an integer from {1, 2,...,16}. Node 8 is the root of the tree. No…

Nearest number - 2 Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 3943 Accepted: 1210 Description Input is the matrix A of N by N non-negative integers.  A distance between two elements Aij and Apq is defined as |i − p| + |j − q|.  Your p…

Nearest number - 2 Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 4100 Accepted: 1275 Description Input is the matrix A of N by N non-negative integers. A distance between two elements Aij and Apq is defined as |i − p| + |j − q|. Your pro…

POJ.1330 Nearest Common Ancestors (LCA 倍增) 题意分析 给出一棵树,树上有n个点(n-1)条边,n-1个父子的边的关系a-b.接下来给出xy,求出xy的lca节点编号. LCA裸题,用倍增思想. 代码总览 #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #define nmax 80520 #define demen…