51nod 1095 Anigram单词

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1095

竟然错了10次,最后还是用字典树搞过去的,不过这题解法很多,二分,哈希,STL,排序 都可以搞。

字典树建树的时候保存节点出现的次数,因为可能大小写都有,所以开next[52]的数组足够了.

题目关键是相同字符串不能算,那么可以用map统计出现次数最后减去这个即可.

还有就是查找的时候,边查找边记数一直到字符串的结尾,并且如果还有分支的话那么需要减去分支的数,因为是不属于当前字符串的.

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>
//#pragma comment(linker, "/STACK:102400000,102400000")
#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define ll long long
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)

#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("a.txt", "r", stdin)
#define Write() freopen("b.txt", "w", stdout);
#define maxn 1010
#define maxv 1010
#define mod 1000000000
using namespace std;

typedef struct node
{
    int count;
    struct node *next[52];
}*tree;

void insert(tree h,string s)
{
    tree p=h;
    int len=s.length();
    for(int i=0;i<len;i++)
    {
        int index=s[i]-‘A‘;
        if(p->next[index]!=NULL)
        {
            p=p->next[index];
            p->count++;
        }
        else
        {
            tree tem=(tree)calloc(1,sizeof(node));
            tem->count=1;
            p->next[index]=tem;
            p=tem;
        }
    }
}

int find(tree h,string s)
{
    tree p=h;
    int len=s.length();
    for(int i=0;i<len;i++)
    {
        int index=s[i]-‘A‘;
        if(p->next[index]==NULL)
            return -1;
        p=p->next[index];
    }
    int ans=p->count;
    for(int i=0;i<52;i++)
        if(p->next[i]!=NULL) ans-=p->next[i]->count;
    //cout<<ans<<endl;
    return ans;
}
int main()
{
    //freopen("a.txt","r",stdin);
    int n,m;
    string s;
    map<string,int>ma;
    cin>>n;
    tree head=(tree)calloc(1,sizeof(node));
    while(n--)
    {
        cin>>s;
        ma[s]++;
        sort(s.begin(),s.end());
        insert(head,s);
    }
    cin>>m;
    while(m--)
    {
        cin>>s;
        int ans=0-ma[s];
        //cout<<ans<<endl;
        sort(s.begin(),s.end());
        int x=find(head,s);
        if(x==-1) cout<<0<<endl;
        else cout<<ans+x<<endl;
    }
    return 0;
}

大神的哈希表代码：

 1 //#pragma comment(linker,"/STACK:102400000,102400000")
 2 #include<stdio.h>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<math.h>
 6 #include<algorithm>
 7 #include<vector>
 8 #include<map>
 9 #include<set>
10 #include<queue>
11 #include<string>
12 #define ll long long
13 #define db double
14 #define PB push_back
15 #define lson k<<1
16 #define rson k<<1|1
17 using namespace std;
18
19 const int N = 10005;
20 const int MOD = 100007;
21 const int base = 57;
22
23 map<string,int> mp;
24 char str[20];
25 int cnt[base+10],h[MOD+10];
26
27 int hs()
28 {
29     memset(cnt,0,sizeof(cnt));
30     for(int i=0;str[i];i++)
31     {
32         if(str[i]>=‘a‘&&str[i]<=‘z‘) cnt[str[i]-‘a‘]++;
33         else cnt[str[i]-‘A‘+26]++;
34     }
35     int res=0;
36     for(int i=0;i<base;i++)
37     {
38         res=(res*base+cnt[i])%MOD;
39     }
40     return res;
41 }
42
43 int main()
44 {
45 #ifdef PKWV
46     freopen("in.in","r",stdin);
47 #endif // PKWV
48     int n;
49     scanf("%d",&n);
50     int lm=1;
51     string s;
52     for(int i=0;i<n;i++)
53     {
54         scanf("%s",str);
55         h[hs()]++;
56         string s=str;
57         if(mp[s]==0) mp[s]=lm++;
58     }
59     int q;
60     scanf("%d",&q);
61     while(q--)
62     {
63         scanf("%s",str);
64         int len=strlen(str);
65         int cnt=0;
66         int res=hs();
67         cnt=h[res];
68         s=str;
69         if(mp[s]>0) cnt--;
70         printf("%d\n",cnt);
71     }
72     return 0;
73 }