u010652938的专栏

#include #include #include #include #include using namespace std;int sum, g[30][30], d[30], en, way[30], l_w;int maxd, flag[30], l_f;int cmp(const void *a, const void *b){return *(int*)a-*(in

因为格式WA了好多次，居然不是PE，受不鸟，还以为哪错了。。。坑爹的JUDGE！#include #include #include #include using namespace std;bool is_prime(int n){ if(n == 1) return false; for(int i = 2; i < n; i++)

#include#include#includeusing namespace std;char word[102][102];char s[102];int stx,sty,enx,eny,N,len;int dire[8][2]= {0,1,0,-1,-1,0,1,0,-1,-1,-1,1,1,-1,1,1};bool DFS(int i,int j,int dir,int p

#include #includeusing namespace std;#define MAXN 15int main(void){ char map[MAXN][MAXN]; int i,j,n,t,p; char ch; scanf("%d",&t); p=1; while(t--) { printf("Cas

#include #includeusing namespace std;bool visit[6][6];int an[30];void solve(int u,int res){ if(res == 9) { for(int i=0; i<res; ++i) cout<<an[i]; cout<<endl

#include #include #include #include #include #include #define maxn 16 + 10#define ll long long#define INF 1000000000#define FOR(i, a, b) for(int i = a; i < b; ++i)using namespace std;int n

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，大熊要从左上角走

#include #include #include #include #include #include #define maxn 5 + 10#define ll long long#define INF 1000000000#define FOR(i, a, b) for(int i = a; i < b; ++i)using namespace std;bool c

#include #include #include #include #include using namespace std;void dfs(char s[210][210], int x, int y, int dep){ for(int i = y; s[x][i] == '-'; ++i) if(s[x+1][i] != ' ' && s[x+1

#include #include #include #include using namespace std;int ans[30][30], vis[30][30];int len, n, m;void dfs(int v, int l){ if(len < l) len = l; for(int i = 0; i < n; i++) {

wa了两次，就是少了负数和0（）的情况。。。。#include #include #include #include #include #include #include using namespace std;stack s;stack num;int n, flag, ok;char ch;void dfs(int sum){ if(ch == '(')

虽然说AC了，但是还是不太懂，参考了他人的代码，有点烦，仔细点其实不算太难。#include #include #include using namespace std;struct room {int energy,doorways[120],num_doorways,last_energy;};room r[120];int n;bool is_vis[120];bool

#include #include #include #include #include using namespace std;struct dot{ int x[50], y[50], z, num;} d[1000][1000];int vis[1000][1000];int dfs(int x, int y){ if(!d[x][y].z) retur

用1， 0表示，/ 为001  \ 表示为100， 然后找回路；                            010                010                            100                001#include #include #include #include #include using std:

#include #include #include #include #include using namespace std;char grid[33][88];int h, move[4][2]= {0,1,0,-1,1,0,-1,0};void dfs(int x, int y){ //cout << x << " " << y << endl; if(g

dfs嵌套。#include #include #include #include #include using namespace std;char board[55][55];int move[4][2]= {0,-1,0,1,-1,0,1,0};int vis[55][55], num[50], len, sum, w, h;void ddfs(int x, int

这个R有可能为零，为零的时候各数据也都OK，需要注意。#include#include#include#include using namespace std;int N, R, ok;int head[10010], end[10010], vis[10010];int degree[210], flag, num;void dfs(int cur){ for(

八皇后计数，求最大。#include#include#include#include using std::cin;using std::cout;using std::endl;int board[10][10], vis[3][20], c[10], n, tot;void dfs(int cur){ if(cur == 8) { int

欧拉函数算n以内的质数。#include #include #include #include #include #include using namespace std;int vis[50010], sum[50010];void ol(){ memset(vis, 0, sizeof(vis)); sum[0] = 0; vis[1] = 1;

#include #include #include #include #include #include #include #include #include #include #include using namespace std;#define maxn 100 + 10#define INF 2100000000#define ll long longboo