The Preliminary Contest for ICPC Asia Xuzhou 2019 G. Colorful String 回文树

签到提；

题意：求出每一个回文串的贡献 (贡献的计算就是回文串不同字符的个数)

题解：

用回文树直接暴力即可

回文树开一个数组cost[ ][26] 和val[ ] 数组；

val【i】表示回文树上节点 i 的对应的回文的贡献

最后统计答案即可

LL get_ans() {
    LL ans = 0;
    for (int i = sz - 1; i >= 0; --i) ans += 1LL * cnt[i] * val[i];
    return ans;
}

 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <unordered_map>

 #define  pi    acos(-1.0)
 #define  eps   1e-9
 #define  fi    first
 #define  se    second
 #define  rtl   rt<<1
 #define  rtr   rt<<1|1
 #define  bug                printf("******\n")
 #define  mem(a, b)          memset(a,b,sizeof(a))
 #define  name2str(x)        #x
 #define  fuck(x)            cout<<#x" = "<<x<<endl
 #define  sfi(a)             scanf("%d", &a)
 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  sfL(a)             scanf("%lld", &a)
 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 #define  sfs(a)             scanf("%s", a)
 #define  sffs(a, b)         scanf("%s %s", a, b)
 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 #define  FIN                freopen("../in.txt","r",stdin)
 #define  gcd(a, b)          __gcd(a,b)
 #define  lowbit(x)          x&-x
 #define  IO                 iOS::sync_with_stdio(false)

 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 const ULL seed = ;
 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 const int maxn = 4e5 + ;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = 1e9 + ;
 char s[maxn];

 struct Palindrome_Automaton {
     int len[maxn], next[maxn][], fail[maxn], cnt[maxn], cost[maxn][], val[maxn];
     int num[maxn], S[maxn], sz, n, last;

     int newnode(int l) {
         for (int i = ; i < ; ++i)next[sz][i] = , cost[sz][i] = ;
         cnt[sz] = num[sz] = , len[sz] = l;
         return sz++;
     }

     void init() {
         sz = n = last = ;
         newnode();
         newnode(-);
         S[] = -;
         fail[] = ;
     }

     int get_fail(int x) {
         while (S[n - len[x] - ] != S[n])x = fail[x];
         return x;
     }

     void add(int c, int pos) {
         c -= 'a';
         S[++n] = c;
         int cur = get_fail(last);
         if (!next[cur][c]) {
             int now = newnode(len[cur] + );
             fail[now] = next[get_fail(fail[cur])][c];
             next[cur][c] = now;
             num[now] = num[fail[now]] + ;

             for (int i = ; i < ; i++) cost[now][i] = cost[cur][i];
             cost[now][c] = ;
             int temp = ;
             for (int i = ; i < ; i++) temp += cost[now][i];
             val[now] = temp;

         }
         last = next[cur][c];
         cnt[last]++;
     }

     void count()//统计本质相同的回文串的出现次数
     {
         for (int i = sz - ; i >= ; --i)cnt[fail[i]] += cnt[i];
         //逆序累加，保证每个点都会比它的父亲节点先算完，于是父亲节点能加到所有子孙
     }

     LL get_ans() {
         LL ans = ;
         for (int i = sz - ; i >= ; --i) ans += 1LL * cnt[i] * val[i];
         return ans;
     }
 } pam;

 int main() {
     FIN;
     sfs(s + );
     pam.init();
     int n = strlen(s + );
     for (int i = ; i <= n; i++) {
         pam.add(s[i], i);
     }
     pam.count();
     LL ans = pam.get_ans();
     printf("%lld\n", ans);
     return ;
 }