The Sultan's Successors Description The Sultan of Nubia has no children, so she has decided that the country will be split into up to k separate parts on her death and each part will be inherited by whoever performs best at some test. It is possible…

  题意:八皇后问题的扩展.8*8棋盘上每个格子都有一个整数,要求8个皇后所在格子的数字之后最大 解法一,回溯: 用vis数组记录 列,主对角(y-x), 副对角(y+x) 访问情况 #include<cstdio> #include<cstring> #include<iostream> #include<string> #include<algorithm> using namespace std; ], vis[][], tot = , n…

https://vjudge.net/contest/68264#problem/R The Sultan of Nubia has no children, so she has decided that the country will be split into up to k separate parts on her death and each part will be inherited by whoever performs best at some test. It is po…

the squares thus selected sum to a number at least as high as one already chosen by the Sultan. (For those unfamiliar with the rules of chess, this implies that each row and column of the board contains exactly one queen, and each diagonal contains n…

题目链接:Uva 167 思路分析:八皇后问题,采用回溯法解决问题. 代码如下: #include <iostream> #include <string.h> using namespace std; ; int A[MAX_N]; int M[MAX_N][MAX_N]; ; int is_safe( int row, int col ) { ; i < row; ++i ) { if ( A[i] == col ) return false; if ( A[i] + i…

※Recorded By ksq2013 //其实这段时间写的题远远大于这篇博文中的内容,只不过那些数以百记的基础题目实在没必要写在blog上; ※week one 2016.7.18 Monday a)bzoj4034[HAOI2015 T2](树链剖分+线段树) http://blog.csdn.net/keshuqi/article/details/51944955 b)luogu1328[NOIP2014 T1]生活大爆炸版石头剪刀布(模拟) http://blog.csdn.net/k…

题目例如以下: The Sultan's Successors  The Sultan of Nubia has no children, so she has decided that thecountry will be split into up to k separate parts on her death andeach part will be inherited by whoever performs best at some test. Itis possible for an…

题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined Surfaces 490 - Rotating Sentences 445 - Marvelous Mazes…

这道题是典型的八皇后问题,刘汝佳书上有具体的解说. 代码的实现例如以下: #include <stdio.h> #include <string.h> #include <stdlib.h> int vis[100][100];//刚開始wrong的原因就是这里数组开小了,开了[8][8],以为可以.后来看到[cur-i+8]意识到可能数组开小了.改大之后AC. int a[8][8]; int C[10]; int max_,tot; void search_(int…

题目大意: 有若干模式串,将某些模式串拼接起来(一个可以使用多次)形成一个长模式串,判断能否有两种或更多种不同的拼法拼成相同的模式串. 思路: 神奇的构图,暴力的求解. 可以发现,若有不同的拼法,则一个模式串的前缀要与一个模式串的后缀相同. 因此我们就将问题转化成:从两个模式串开始,不停的按照前后缀匹配,最后达到两个串同时在一个点结束. 那么,将每一个串的每一个字符都看作一个点,n2len2暴力枚举i串从z开始的后缀和j串(自己也可以,但不能让前缀是其本身)的前缀做匹配,看是否能将其中一个串匹配…