2015
07-16

# Palindromic Substring

In the kingdom of string, people like palindromic strings very much. They like only palindromic strings and dislike all other strings. There is a unified formula to calculate the score of a palindromic string. The score is calculated by applying the following three steps.
1. Since a palindromic string is symmetric, the second half (excluding the middle of the string if the length is odd) is got rid of, and only the rest is considered. For example, "abba" becomes "ab", "aba" becomes "ab" and "abacaba" becomes "abac".
2. Define some integer values for ‘a’ to ‘z’.
3. Treat the rest part as a 26-based number M and the score is M modulo 777,777,777.
However, different person may have different values for ‘a’ to ‘z’. For example, if ‘a’ is defined as 3, ‘b’ is defined as 1 and c is defined as 4, then the string "accbcca" has the score (3×263+4×262+4×26+1) modulo 777777777=55537.
One day, a very long string S is discovered and everyone in the kingdom wants to know that among all the palindromic substrings of S, what the one with the K-th smallest score is.

The first line contains an integer T(1 ≤ T ≤ 20), the number of test cases.
The first line in each case contains two integers n, m (1 ≤ n ≤ 100000, 1 ≤ m ≤ 20) where n is the length of S and m is the number of people in the kingdom. The second line is the string S consisting of only lowercase letters. The next m lines each containing 27 integers describes a person in the following format.
Ki va vb … vz

Where va is the value of ‘a’ for the person, vb is the value of ‘b’ and so on. It is ensured that the Ki-th smallest palindromic substring exists and va, vb, …, vz are in the range of [0, 26). But the values may coincide.

The first line contains an integer T(1 ≤ T ≤ 20), the number of test cases.
The first line in each case contains two integers n, m (1 ≤ n ≤ 100000, 1 ≤ m ≤ 20) where n is the length of S and m is the number of people in the kingdom. The second line is the string S consisting of only lowercase letters. The next m lines each containing 27 integers describes a person in the following format.
Ki va vb ... vz

Where va is the value of 'a' for the person, vb is the value of 'b' and so on. It is ensured that the Ki-th smallest palindromic substring exists and va, vb, ..., vz are in the range of [0, 26). But the values may coincide.

3
6 2
abcdca
3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
7 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
4 10
zzzz
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
7 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
9 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
10 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14
51 4
abcdefghijklmnopqrstuvwxyzyxwvutsrqponmlkjihgfedcba
1 1 3 3 25 20 25 21 7 0 9 7 3 16 15 14 19 5 19 19 19 22 8 23 2 4 1
25 1 3 3 25 20 25 21 7 0 9 7 3 16 15 14 19 5 19 19 19 22 8 23 2 4 1
26 1 3 3 25 20 25 21 7 0 9 7 3 16 15 14 19 5 19 19 19 22 8 23 2 4 1
76 1 3 3 25 20 25 21 7 0 9 7 3 16 15 14 19 5 19 19 19 22 8 23 2 4 1

1
620

14
14
14
14
14
14
14
378
378
378

0
9
14
733665286

Hint
There are 7 palindromic substrings {"a", "a", "b", "c", "c", "d", "cdc"} in the first case. For
the first person, the corresponding scores are {1, 1, 1, 1, 1, 1, 27}. For the second person, the corresponding scores are {25, 25, 24, 23, 23, 22, 620}.


HDU 判定结果:

ZOJ 判定结果:

/*
* Author: Gatevin
* Created Time:  2015/3/26 19:01:01
* File Name: Chitoge_Kirisaki.cpp
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;
typedef unsigned long long ulint;
#define maxn 100100
#define rank rrank
const ulint seed = 50009;
const lint mod = 777777777LL;
set <ulint> S;
ulint H[maxn], xp[maxn];
lint W[maxn], xp2[maxn], w[30], K, cnt[maxn];
char in[maxn], in_new[maxn << 1];
int s[maxn], sa[maxn], R[maxn << 1];
vector<pair<lint, lint> > result;

void inithash(int n)//对于原字符串的hash判断回文串种类
{
H[0] = in[0] – 'a' + 1;
for(int i = 1; i < n; i++)
H[i] = H[i - 1]*seed + (ulint)(in[i] – 'a' + 1);
return;
}

{
if(l == 0) return H[r];
return H[r] – H[l - 1]*xp[r - l + 1];
}

void initWeight(int n)
{
W[0] = w[in[0] – 'a'];
for(int i = 1; i < n; i++)
W[i] = (W[i - 1]*26LL + w[in[i] – 'a']) % mod;
return;
}

{
if(l == 0) return W[r];
return (W[r] – W[l - 1]*xp2[r - l + 1] % mod + mod) % mod;
}

int wa[maxn], wb[maxn], wv[maxn], Ws[maxn];

int cmp(int *r, int a, int b, int l)
{
return r[a] == r[b] && r[a + l] == r[b + l];
}

void da(int *r, int *sa, int n, int m)
{
int *x = wa, *y = wb, *t, i, j, p;
for(i = 0; i < m; i++) Ws[i] = 0;
for(i = 0; i < n; i++) Ws[x[i] = r[i]]++;
for(i = 1; i < m; i++) Ws[i] += Ws[i - 1];
for(i = n – 1; i >= 0; i–) sa[--Ws[x[i]]] = i;
for(j = 1, p = 1; p < n; j *= 2, m = p)
{
for(p = 0, i = n – j; i < n; i++) y[p++] = i;
for(i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i] – j;
for(i = 0; i < n; i++) wv[i] = x[y[i]];
for(i = 0; i < m; i++) Ws[i] = 0;
for(i = 0; i < n; i++) Ws[wv[i]]++;
for(i = 1; i < m; i++) Ws[i] += Ws[i - 1];
for(i = n – 1; i >= 0; i–) sa[--Ws[wv[i]]] = y[i];
for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p – 1 : p++;
}
return;
}

int rank[maxn], height[maxn];
void calheight(int *r, int *sa, int n)
{
int i, j, k = 0;
for(i = 1; i <= n; i++) rank[sa[i]] = i;
for(i = 0; i < n; height[rank[i++]] = k)
for(k ? k– : 0, j = sa[rank[i] – 1]; r[i + k] == r[j + k]; k++);
return;
}

int dp[maxn][20];
void initRMQ(int n)
{
for(int i = 1; i <= n; i++) dp[i][0] = height[i];
for(int j = 1; (1 << j) <= n; j++)
for(int i = 1; i + (1 << j) – 1 <= n; i++)
dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
return;
}

{
//int ra = rank[a], rb = rank[b];
int ra = a, rb = b;
if(ra > rb) swap(ra, rb);
int k = 0;
while((1 << (k + 1)) <= rb – ra + 1) k++;
return min(dp[ra][k], dp[rb - (1 << k) + 1][k]);
}

lint calCnt(int l, int r, int n)
{
int rl = rank[l];
int L = rl + 1, R = n, mid;
lint lmost = rl, rmost = rl;
while(L <= R)//向左和向右都二分一次找到左右界
{
mid = (L + R) >> 1;
if(askRMQ(rl + 1, mid) >= (r – l + 1))
{
rmost = mid;
L = mid + 1;
}
else
R = mid – 1;
}
L = 1; R = rl – 1;
while(L <= R)
{
mid = (L + R) >> 1;
if(askRMQ(mid + 1, rl) >= (r – l + 1))
{
lmost = mid;
R = mid – 1;
}
else
L = mid + 1;
}
return rmost – lmost + 1;
}

vector <pair<int, int> > pal;

void Manacher(char *s, int *R, int n)
{
S.clear();
int p = 0, mx = 0;
R[0] = 1;
for(int i = 1; i < n; i++)//第n个字符不要算..没注意这里WA了好几次….
{
if(mx > i) R[i] = min(R[2*p - i], mx – i);
else R[i] = 1;
while(s[i - R[i]] == s[i + R[i]])
R[i]++;
if(i + R[i] > mx)
{
for(int j = mx; j < i + R[i]; j++)//每一次mx的位移都可能是一个新的回文串
{
int l = 2*i – j, r = j;
l >>= 1;
r = (r & 1) ? r >> 1 : (r >> 1) – 1;//对应的回文串的原位置[l, r]
if(l > r) continue;
set<ulint> :: iterator it = S.find(hashvalue);
if(it == S.end())
{
S.insert(hashvalue);
pal.push_back(make_pair(l, r));
}
}
mx = i + R[i], p = i;
}
}
return;
}

int main()
{
int T;
scanf("%d", &T);
xp[0] = 1, xp2[0] = 1;
for(int i = 1; i < maxn; i++)
xp[i] = xp[i - 1]*seed, xp2[i] = xp2[i - 1]*26LL % mod;
while(T–)
{
int n, m;
scanf("%d %d", &n, &m);
scanf("%s", in);
in_new[0] = '@';
s[0] = in[0] – 'a' + 1;
for(int i = 0; i < n; i++)
in_new[2*i + 1] = in[i], in_new[2*i + 2] = '#', s[i] = in[i] – 'a' + 1;
in_new[2*n] = '\$';
s[n] = 0;
da(s, sa, n + 1, 28);
calheight(s, sa, n);
initRMQ(n);
pal.clear();
inithash(n);
Manacher(in_new, R, 2*n);
for(int i = pal.size() – 1; i >= 0; i–)//后缀数组二分查找每种回文串的数量
cnt[i] = calCnt(pal[i].first, pal[i].second, n);
while(m–)
{
scanf("%I64d", &K);
for(int i = 0; i < 26; i++)
scanf("%I64d", w + i);
initWeight(n);
result.clear();
for(int i = pal.size() – 1; i >= 0; i–)//对第i种回文串类似hash的方法求出权值
{
lint weight = askWeight(pal[i].first, (pal[i].first + pal[i].second) >> 1);
result.push_back(make_pair(weight, cnt[i]));
}
sort(result.begin(), result.end());
int now = 0;
while(K > result[now].second)
K -= result[now].second, now++;
printf("%I64d\n", result[now].first);
}
printf("\n");
}
return 0;
}


1. 第二个点球多少有点勉强，第三个进球精彩，郜林传于大宝那个球，虽然没进，的确是国足少见的配合。裁判澳大利亚的，有点中国裁判的水准