CSU 1719（智障选手的DFS）
1719: Boggle
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 253 Solved: 28
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Description
Input
Output
Sample Input
3
april
purple
quilt
5
rprit
ahqln
ietep
zrysg
ogwey
3
pel
aup
bcr
0
Sample Output
april
quilt
-
purple
-
HINT
Source
题意：有n个单词，然后给你一个m*m的字符矩阵，让你找出一条路径使得走过的路径是一个单词，方向是八个方向都可以走。问有哪些单词可以匹配？这里如果字符矩阵里出现了q，那么把他看作是qu
题解：智障选手写了好久的题，本来很简单的暴力DFS，可是忘记加标记访问。
这里给出2份代码，第一份直接爆搜，第二份是将单词中所有的qu转化为q，这样搜起来更简单一点
第一份：直接爆搜
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<bitset>
#include<utility>
#include<functional>
#include<iomanip>
#include<sstream>
#include<ctime>
using namespace std;
#define N int(3e3)
#define inf int(0x3f3f3f3f)
#define mod int(1e9+7)
typedef long long LL;
int vx[] = { 0,1,0,-1,1,-1,1,-1 };
int vy[] = { 1,0,-1,0,-1,1,1,-1 };
char s[N][N];
char g[N][N];
int vis[N][N];
int t, n;
bool judge(int x, int y)
{
if (x<0 || x >= n || y<0 || y >= n)
return 0;
if (vis[x][y])return 0;
return 1;
}
int ok = 0;
void dfs(int x, int y, int dep, int num, int length)
{
if (dep == length)
{
ok = 1;
return;
}
if (ok || dep>length)
{
return;
}
for (int i = 0; i<8; i++)
{
int xx = x + vx[i];
int yy = y + vy[i];
if (judge(xx, yy))
{
if (g[xx][yy] == s[num][dep])
{
vis[xx][yy] = 1;
if (s[num][dep] == 'q'&&s[num][dep + 1] == 'u')
{
dfs(xx, yy, dep + 2, num, length);
}
else if (s[num][dep] != 'q')
{
dfs(xx, yy, dep + 1, num, length);
}
vis[xx][yy] = 0;
}
}
}
}
vector<string>ans;
void solve()
{
ans.clear();
memset(vis, 0, sizeof(vis));
for (int k = 0; k<t; k++)
{
ok = 0;
for (int i = 0; i<n; i++)
{
for (int j = 0; j<n; j++)
{
if (s[k][0] == g[i][j] && s[k][0] == 'q'&&s[k][1] == 'u')
{
vis[i][j]=1;
dfs(i, j, 2, k, strlen(s[k]));
vis[i][j]=0;
}
else if (s[k][0] == g[i][j] && s[k][0] != 'q')
{
vis[i][j]=1;
dfs(i, j, 1, k, strlen(s[k]));
vis[i][j]=0;
}
if (ok)break;
}
if (ok)break;
}
if (ok)
{
ans.push_back(string(s[k]));
}
}
sort(ans.begin(), ans.end());
for (int i = 0; i<ans.size(); i++)
{
printf("%s\n", ans[i].c_str());
}
}
int main()
{
#ifdef CDZSC
freopen("i.txt", "r", stdin);
//freopen("o.txt","w",stdout);
int _time_jc = clock();
#endif
while (~scanf("%d", &t))
{
for (int i = 0; i < t; i++)
{
scanf("%s", s[i]);
}
while (~scanf("%d", &n))
{
if (n == 0)break;
for (int i = 0; i < n; i++)
scanf("%s", g[i]);
solve();
puts("-");
}
}
return 0;
}
第二份：将单词中所有的qu转化为q，这样搜起来更简单一点
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<bitset>
#include<utility>
#include<functional>
#include<iomanip>
#include<sstream>
#include<ctime>
using namespace std;
#define N int(1e3)
#define inf int(0x3f3f3f3f)
#define mod int(1e9+7)
typedef long long LL;
int vx[] = { 0,1,0,-1,1,-1,1,-1 };
int vy[] = { 1,0,-1,0,-1,1,1,-1 };
char s[N][N];
char g[N][N];
int vis[N][N];
int t, n;
map<string, string>st;
bool judge(int x, int y)
{
if (x<0 || x >= n || y<0 || y >= n)
return 0;
if (vis[x][y])return 0;
return 1;
}
int ok = 0;
void dfs(int x, int y, int dep, int num, int length)
{
if (dep == length)
{
ok = 1;
return;
}
if (ok || dep>length)
{
return;
}
for (int i = 0; i<8; i++)
{
int xx = x + vx[i];
int yy = y + vy[i];
if (judge(xx, yy))
{
if (g[xx][yy] == s[num][dep])
{
vis[xx][yy] = 1;
dfs(xx, yy, dep + 1, num, length);
vis[xx][yy] = 0;
}
}
}
}
vector<string>ans;
void solve()
{
ans.clear();
memset(vis, 0, sizeof(vis));
for (int k = 0; k<t; k++)
{
ok = 0;
for (int i = 0; i<n; i++)
{
for (int j = 0; j<n; j++)
{
if (s[k][0] == g[i][j])
{
vis[i][j] = 1;
dfs(i, j, 1, k, strlen(s[k]));
vis[i][j] = 0;
}
if (ok)break;
}
if (ok)break;
}
if (ok)
{
ans.push_back(st[string(s[k])]);
}
}
sort(ans.begin(), ans.end());
for (int i = 0; i<ans.size(); i++)
{
printf("%s\n", ans[i].c_str());
}
}
int main()
{
#ifdef CDZSC
freopen("i.txt", "r", stdin);
//freopen("o.txt","w",stdout);
int _time_jc = clock();
#endif
scanf("%d", &t);
for (int i = 0; i<t; i++)
{
string tmp;
scanf("%s", s[i]);
for (int j = 0; s[i][j]; j++)
{
if (s[i][j] == 'q'&&s[i][j + 1] == 'u')
{
j++;
tmp += "q";
}
else if (s[i][j] == 'q')
{
tmp += "#";
}
else
{
tmp += s[i][j];
}
}
st[tmp] = string(s[i]);
for (int j = 0; j<tmp.size(); j++)
{
s[i][j] = tmp[j];
}
s[i][tmp.size()] = '\0';
}
while (~scanf("%d", &n) && n)
{
for (int i = 0; i<n; i++)
scanf("%s", g[i]);
solve();
puts("-");
}
return 0;
}