gzdrn的博客

简单的搜索，多一个维度#include#includeint l,r,c;int sx,sy,sz;char a[31][31][31];int book[31][31][31];int xx[6]={1,-1,0,0,0,0};int yy[6]={0,0,1,-1,0,0};int zz[6]={0,0,0,0,1,-1};struct note{ int x,y

Find The MultipleTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 24082Accepted: 9981Special JudgeDescriptionGiven a positive integer n, write a

Description定义一个二维数组： int maze[5][5] = { 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0,};它表示一个迷宫，其中的1表示墙壁，0表示可以走的路，只能横着走或竖着走，不能斜着走，要求编程序找出从左上角到右下角的最短

大意为@相邻的八个位置的@都为同一子块，求子块的个数。依稀记得学过floodfill的方法求子块个数，但是忘了。。。于是看了一下题解，发现可以遍历图时DFS实现计数#include#includechar a[101][101];int book[101][101];int m,n;int movex[8]={1,1,1,-1,-1,-1,0,0},movey[8]={1,0

从可行性出发容易想到BFS,但是每种状态下有3种情况拓展（i-1,i+1,2*i）数目有点大，因此需要剪枝1.每个数最大取到k+1才有意义2.每个数出现一次即可，因此可以用一下hash技术a[]数组#includeint n,k;struct note{ int num; int s;}que[1000000];int a[1000000];void bfs(voi

感觉还是不熟，忙了好久。#include#includeint n,m;char map[205][205];int a[205][205],b[205][205];int book[205][205];int x1,y1,x2,y2;struct note{ int x,y; int s;}que[100000];int next[4][2]={{0,1},{

DescriptionAs every one known, a football team has 11 players . Now, there is a big problem in front of the Coach Liu. The final contest is getting closer. Who is the center defense, the full back

#include#include#include#includeusing namespace std;int n;int a[21]={0};int book[21]={0};int prime[100];void dfs(int count)//count 为已排好的个数,回溯法{ int i,j; int temp; if(count==n) { for(i