状态压缩+枚举 UVA 11464 Even Parity】的更多相关文章

题目传送门 /* 题意:求最少改变多少个0成1,使得每一个元素四周的和为偶数 状态压缩+枚举:枚举第一行的所有可能(1<<n),下一行完全能够由上一行递推出来,b数组保存该位置需要填什么 最后检查不同的数量,取最小值 */ #include <cstdio> #include <algorithm> #include <cstring> using namespace std; ; const int INF = 0x3f3f3f3f; int a[MAXN…
UVA 1508 - Equipment 状态压缩 枚举子集 dfs ACM 题目地址:option=com_onlinejudge&Itemid=8&category=457&page=show_problem&problem=4254" target="_blank" style="color:rgb(0,136,204); text-decoration:none">UVA 1508 - Equipment--P…
题目链接:UVA 811 Description Once upon a time, in a faraway land, there lived a king. This king owned a small collection of rare and valuable trees, which had been gathered by his ancestors on their travels. To protect his trees from thieves, the king or…
UVA.11464 Even Parity (思维题 开关问题) 题目大意 给出一个n*n的01方格,现在要求将其中的一些0转换为1,使得每个方格的上下左右格子的数字和为偶数(如果存在的话),求使得最小的转换的个数. 最先想到的是枚举每个方格的状态,非0即1,那么就有2^(n*n)种情况,可见数量之大,必定超时.那么就必须要转换思路. 不难看出这是一个开关问题,就是说可以根据一行的数据,推算出下一行的数据,然后顺次推算出整个格子的数据,之后再来看改变了多少的01序列,求出最小的结果.(稍后会演示…
B. Preparing Olympiad You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made. A problemset for the contest must consist of at le…
/* 看别人的的思路 搜索搜不出来我太挫了 状态压缩枚举+好的位置 */ #include<stdio.h> #include<string.h> #define N 20 int lower[N]; char s[N]; int len; int judge(int deng,int l,int r) { int i,f1,f2; int sum1,sum2; f1=s[0]-'0'; sum1=0; for(i=1;i<deng;i++) {//根据状态求出左边的所有和…
题目传送门 /* 题意:问最少翻转几次使得棋子都变白,输出翻转的位置 状态压缩+枚举:和之前UVA_11464差不多,枚举第一行,可以从上一行的状态知道当前是否必须翻转 */ #include <cstdio> #include <cstring> #include <algorithm> using namespace std; ; const int INF = 0x3f3f3f3f; int a[MAXN][MAXN]; int b[MAXN][MAXN]; in…
题目链接:https://www.luogu.com.cn/problem/P1036 题目描述 已知 \(n\) 个整数 \(x_1,x_2,-,x_n\) ,以及 \(1\) 个整数 \(k(k<n)\) . 从 \(n\) 个整数中任选 \(k\) 个整数相加,可分别得到一系列的和. 例如当 \(n=4,k=3\) , \(4\) 个整数分别为 \(3,7,12,19\) 时,可得全部的组合与它们的和为: \(3+7+12=22\) \(3+7+19=29\) \(7+12+19=38\)…
11464 - Even Parity Time limit: 3.000 seconds We have a grid of size N x N. Each cell of the grid initially contains a zero(0) or a one(1).  The parity of a cell is the number of 1s surrounding that cell. A cell is surrounded by at most 4 cells (top,…
Strange Billboard 题目链接: http://acm.hust.edu.cn/vjudge/contest/129733#problem/A Description The marketing and public-relations department of the Czech Technical University has designed a new reconfigurable mechanical Flip-Flop Bill-Board (FFBB). The b…