UVALive - 5913 Dictionary Size

题意：求有前后缀组成的不同的字符串

思路：先用Trie树来统计前后缀不同的字符串，然后就是去重

例如：前后缀分别是x1x1x1x1a,ax2x2x2x2x2的话，那么可能重复的字符串

x1x1x1x1ax2x2x2x2x2,统计前后缀分别以a结尾的个数，然后减去

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 500010;
const int M = 26;

char s[10010][45];
int num1[26],num2[26],C[26],id[MAXN],sz;
int ch[MAXN][M];

void init(){
    sz = 1;
    memset(ch[0],0,sizeof(ch[0]));
    for (int i = 0; i < M; i++)
        id[i+'a'] = i;
}

void insert(char *s,int num[],int &cnt){
    int u = 0,first = 1;
    int len = strlen(s);
    for (int i = 0; i < len; i++){
        int c = id[s[i]];
        if (!ch[u][c]){
            memset(ch[sz],0,sizeof(ch[sz]));
            cnt++;
            if (!first)
                num[c]++;
            ch[u][c] = sz++;
        }
        first = 0;
        u = ch[u][c];
    }
}

int main(){
    int n;
    while (scanf("%d",&n) != EOF){
        memset(num1,0,sizeof(num1));
        memset(num2,0,sizeof(num2));
        memset(C,0,sizeof(C));
        init();
        int cnt1 = 0,cnt2 = 0;
        for (int i = 0; i < n; i++){
            scanf("%s",s[i]);
            insert(s[i],num1,cnt1);
        }
        init();
        for (int i = 0; i < n; i++){
           int len = strlen(s[i]);
           if (len == 1)
               C[s[i][0]-'a'] = 1;
           std::reverse(s[i],s[i]+len);
           insert(s[i],num2,cnt2);
        }
        long long ans = (long long)cnt1*cnt2;
        for (int i = 0; i < 26; i++){
            if (C[i])
                ans++;
            ans -= (long long)num1[i] * num2[i];
        }
        printf("%lld\n",ans);
    }
    return 0;
}