FOJ 2013 11 月赛 F

Problem 2137 奇异字符串

Accept: 33    Submit: 149

Time Limit: 1000 mSec    Memory Limit : 32768 KB

我就是个傻逼，虽然这题超时了，但我的时间真的向水一样的丢在上面！

 Problem Description

seen喜欢一种特殊的字符串，seen称这种字符串为奇异字符串。奇异字符串可以表示为AxA这种形式，A为一个任意非空字符串，只包含小写字母，x为一个不在A中出现过的小写字母。seen认为一个长度为d的奇异字符串的价值为d*d，不是奇异字符串的字符串没有价值。现给一个只包含小写字母的字符串，统计其所有子串的价值总和。一个字符串的子串是指其中连续的一段字符构成的字符串。这里相同的子串如果在原串中出现的位置不同则视为不同，需要分别进行统计。

 Input

第一行一个整数T，表示有T组数据。

 

每组数据输入一行只包含小写字母的非空字符串，长度不超过100000。

 Output

对于每组数据输出一行表示所有子串的价值总和。

 Sample Input

1

abcdabc

 Sample Output

49

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 100005
#define INF 0x7fffffff
#define ll long long
ll n,m;
ll ans;
vector<int>g[27];
char s[maxn];
ll sa[maxn], t1[maxn], t2[maxn], c[maxn];
ll rnk[maxn], height[maxn], d[maxn][21],v[maxn];
void RMQ_init(){
    for (int i = 0; i < n; i++)d[i][0] = height[i];
    for (int j = 1; (1 << j) <= n;j++)
    for (int i = 0; i + (1 << j) - 1 < n; i++)
        d[i][j] = min(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);
}
int lcp(ll a, ll b){
    a = rnk[a]; b = rnk[b];
    if (a > b)swap(a, b);
    a++;
    int t = v[b - a + 1];
    return min(d[a][t], d[b-(1<<t)+1][t]);
}
void getheight(){
    int i, j, k = 0;
    for (i = 0; i < n; i++)rnk[sa[i]] = i;
    for (i = 0; i < n; i++){
        if (k)k--;
        if (rnk[i] - 1 < 0)continue;
        int j = sa[rnk[i]-1];
        while (s[i + k] == s[j + k])k++;
        height[rnk[i]] = k;
    }
}
void suffix(int m){
    ll  *x = t1, *y = t2;
    int i;
    for (i = 0; i < m; i++)c[i] = 0;
    for (i = 0; i < n; i++)c[x[i] = s[i]-'a']++;
    for (i = 1; i < m; i++)c[i] += c[i - 1];
    for (i = n - 1; i >= 0; i--)sa[--c[x[i]]] = i;
    for (int k = 1; k <= n; ){
        int p = 0;
        for (i = n - k; i < n; i++)y[p++] = i;
        for (i = 0; i < n; i++)if (sa[i] >= k)y[p++] = sa[i] - k;
        for (i = 0; i < m; i++)c[i] = 0;
        for (i = 0; i < n; i++)c[x[y[i]]]++;
        for (i = 1; i < m; i++)c[i] += c[i - 1]; 
        for (i = n - 1; i >= 0; i--)sa[--c[x[y[i]]]] = y[i];
        swap(x, y);
        p = 1; x[sa[0]] = 0;
        for (i = 1; i < n; i++)x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k] ? p - 1 : p++;
        if (p >= n)break;
        m = p;
        k <<= 1;
    }
    getheight();
}
ll fun(ll a, ll b){
    int x = lcp(a, b);
    if (x!=INF&&x>= b - a - 1)return ((b-a)*2-1)*((b-a)*2-1);
    return 0;
}
int main(){
    int t;
    for (int i = 0; i < maxn; i++)
        v[i] = (i&(i - 1)) ? v[i - 1] : v[i - 1] + 1;
    for (int i = 0; i < maxn; i++)v[i] -= 2;
    v[1] = 0; v[2] = v[3] = 1;
    scanf("%d", &t);
    while (t--){
        for (int i = 0; i < 26; i++)g[i].clear();
        scanf("%s", s);
        n = strlen(s);
        m = 0;
        ans = 0;
        v[0] = -1;
        for (int i = 0; i < n; i++)for (int j = 0; i+(1<<j) <= n; j++)d[i][j] = INF;
        suffix(26);
        RMQ_init();
        for (int i = 0; i < n; i++)g[s[i] - 'a'].push_back(i);
        for (int i = 0; i < 26; i++){
            int d=0;
            for (int j = 0; j < g[i].size(); j++){
                if (j == 0 && j == g[i].size() - 1)d = min((ll)g[i][j], n - 1 - g[i][j]);
                else{
                    if (j == 0)d = min(g[i][j], g[i][j + 1] - g[i][j]-1);
                    else if (j == g[i].size() - 1)d = min((ll)g[i][j] - g[i][j - 1]-1, n - 1 - g[i][j]);
                    else d = min(g[i][j] - g[i][j - 1]-1, g[i][j + 1] - g[i][j]-1);
                }
                for (int k = d; k >= 1; k--){
                    ll x = 0;
                    if (g[i][j] + 1<n)x = fun(g[i][j] - k, g[i][j] + 1);
                    if (x > 0){
                        ans += x;
                        k--;
                    }
                }
            }
        }
        printf("%I64d\n", ans);
    }
}