字典树水题几枚
1.HDU 1251 统计难题
很裸的一道字典树，直接输出个数的， 其实用map也能水过，只不过效率有些低
map版本
map做法就是把每个字符串的所有前缀遍历一遍，然后存起来，最后直接数个数就行。
具体代码吐下 很久很久以前的代码 很挫
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
char s[11];
int main()
{
map<string,int>mp;
int i;
while(gets(s))
{
int len=strlen(s);
if(!len)
break;
string a="";
for(i=0;i<len;i++)
{
a=a+s[i];
mp[a]++;
}
}
char ss[11];
while(scanf("%s",ss)!=EOF)
{
string tmp=ss;
printf("%d\n",mp[tmp]);
}
return 0;
}
/*
ID: sdj22251
PROG: calfflac
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define INF 100000000
#define PI acos(-1.0)
using namespace std;
char str[12];
struct trie
{
int num;
trie *next[26];
trie()
{
num = 0;
for(int i = 0; i < 26; i++)
next[i] = 0;
}
} *root, temp[500005]; // 为了避免用new太多浪费时间，事先开好一个数组， 当然，内存又有可能爆掉了
int ncount = 0;
void insert(trie *p, char *s)
{
int ix = 0;
p ->num++;
while(s[ix])
{
int nx = s[ix] - 'a';
if(p -> next[nx])
{
p = p -> next[nx];
ix++;
}
else
{
p -> next[nx] = &temp[ncount++]; //直接取址
p = p -> next[nx];
ix++;
}
p -> num++;
}
}
int find(trie *p, char *s)
{
int ix = 0, ans;
while(s[ix])
{
int nx = s[ix] - 'a';
if(p -> next[nx])
{
p = p -> next[nx];
ans = p -> num;
ix++;
}
else return 0;
}
return ans;
}
int main()
{
//freopen("ride.in","r",stdin);
//freopen("ride.out","w",stdout);
root = new trie;
while(gets(str))
{
if(strlen(str) == 0) break;
insert(root, str);
}
while(scanf("%s", str) != EOF)
{
printf("%d\n", find(root, str));
}
return 0;
}
2.POJ 2001 Shortest Prefixes
同样，很裸的字典树。
/*
ID: sdj22251
PROG: calfflac
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAX 100000000
#define PI acos(-1.0)
using namespace std;
char s[1005][23];
struct trie
{
int num;
trie *next[26];
trie()
{
num = 0;
for(int i = 0; i < 26; i++)
next[i] = NULL;
}
}*root, temp[50000];
int ncount = 0;
void insert(trie *p, char *s)
{
int ix = 0;
p -> num++;
while(s[ix])
{
int nx = s[ix] - 'a';
if(p -> next[nx])
{
p = p -> next[nx];
ix++;
}
else
{
p -> next[nx] = &temp[ncount++]; //直接取址
p = p -> next[nx];
ix++;
}
p -> num++;
}
}
int find(trie *p, char *s)
{
int ix = 0, ans;
while(s[ix])
{
int nx = s[ix] - 'a';
if(p -> next[nx])
{
p = p -> next[nx];
ans = p -> num;
ix++;
}
else return 0;
}
return ans;
}
int main()
{
//freopen("ride.in","r",stdin);
//freopen("ride.out","w",stdout);
int cnt = 0, i, j;
root = new trie;
while(scanf("%s", s[cnt++]) != EOF)
{
insert(root, s[cnt - 1]);
}
for(i = 0; i < cnt; i++)
{
int len = strlen(s[i]);
char tp[23];
for(j = 0; j < 21; j++)
tp[j] = '\0';
int ct = 0;
for(j = 0; j < len; j++)
{
tp[ct++] = s[i][j];
if(find(root, tp) == 1)
{
break;
}
}
printf("%s %s\n", s[i], tp);
}
return 0;
}
3. HDU 1247
这个就当做模板吧。。。
字典树应用，先插入所有单词，然后暴力拆分每个单词，看拆分后的两个单词是否都在字典树中出现，注意是确切出现，而不是作为子串出现，所以插入的时候注意下
/*
ID: CUGB-wwj
PROG:
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define INF 100000000
#define PI acos(-1.0)
using namespace std;
char str[50051][50];
struct trie
{
int num;
trie *next[26];
trie()
{
num = 0;
for(int i = 0; i < 26; i++)
next[i] = NULL;
}
} *root ;
int ncount = 0;
void insert(trie *p, char *s)
{
int ix = 0;
while(s[ix])
{
int nx = s[ix] - 'a';
if(p -> next[nx] == NULL)
p -> next[nx] = new trie();
p = p -> next[nx];
ix++;
}
p -> num++;
}
int find(trie *p, char *s)
{
int ix = 0, ans = 0;
while(s[ix])
{
int nx = s[ix] - 'a';
if(p -> next[nx])
{
p = p -> next[nx];
ans = p -> num;
ix++;
}
else return 0;
}
return ans;
}
int main()
{
root = new trie();
while(scanf("%s", str[ncount++]) != EOF)
{
insert(root, str[ncount - 1]);
}
for(int i = 0; i < ncount; i++)
{
int len = strlen(str[i]);
int flag = 0;
for(int j = 1; j < len; j++)
{
char ta[22], tb[22];
memset(ta, '\0', sizeof(ta));
memset(tb, '\0', sizeof(tb));
int cnt1 = 0, cnt2 = 0;
for(int k = 0; k < j; k++)
ta[cnt1++] = str[i][k];
for(int k = j; k < len; k++)
tb[cnt2++] = str[i][k];
if(find(root, ta) && find(root, tb))
{
flag = 1;
break;
}
}
if(flag) puts(str[i]);
}
return 0;
}