loi_王振东

题目>>http://codevs.cn/problem/1226/#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int x,y,z,ans = 1e8,map[105][105];void dfs(int a,int b,int c){ if(m

题目>>https://www.luogu.org/problem/show?pid=1149#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int num[11] = {6,2,5,5,4,5,6,3,7,6},n,ans = 0;int find(int

终于调出来了，18ms 粘个题解>>http://codevs.cn/problem/1268/#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int w,d,v,num[50],a,b = 0;bool ok = 0;v

ps：其实只能跑到14； 引用一张图片： 对于某一个放上皇后的点来说，其对于下几行的影响为其所在的列及其（前后列）*向下的行数；#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;int n,t,ans;void dfs(

由于不太会打bfs，这题用的dfs。通过这道题还是学到了蛮多的东西的。首先，瞎打一气（dfs + 标记走过的点）40分。 代码如下：#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int X[] = {0,1,0,-1};const int Y