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("******
")
#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 = 13331;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 4e5 + 7;
const int maxm = 8e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
char s[maxn];

struct Palindrome_Automaton {
    int len[maxn], next[maxn][26], fail[maxn], cnt[maxn], cost[maxn][26], val[maxn];
    int num[maxn], S[maxn], sz, n, last;

    int newnode(int l) {
        for (int i = 0; i < 26; ++i)next[sz][i] = 0, cost[sz][i] = 0;
        cnt[sz] = num[sz] = 0, len[sz] = l;
        return sz++;
    }

    void init() {
        sz = n = last = 0;
        newnode(0);
        newnode(-1);
        S[0] = -1;
        fail[0] = 1;
    }

    int get_fail(int x) {
        while (S[n - len[x] - 1] != 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] + 2);
            fail[now] = next[get_fail(fail[cur])][c];
            next[cur][c] = now;
            num[now] = num[fail[now]] + 1;

            for (int i = 0; i < 26; i++) cost[now][i] = cost[cur][i];
            cost[now][c] = 1;
            int temp = 0;
            for (int i = 0; i < 26; 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 - 1; i >= 0; --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 = 0;
        for (int i = sz - 1; i >= 0; --i) ans += 1LL * cnt[i] * val[i];
        return ans;
    }
} pam;

int main() {
    FIN;
    sfs(s + 1);
    pam.init();
    int n = strlen(s + 1);
    for (int i = 1; i <= n; i++) {
        pam.add(s[i], i);
    }
    pam.count();
    LL ans = pam.get_ans();
    printf("%lld
", ans);
    return 0;
}