HDU 5677 ztr loves substring

Problem Description

ztr love reserach substring.Today ,he has n string.Now ztr want to konw,can he take out exactly k palindrome from all substring of these n string,and thrn sum of length of these k substring is L.

for example string "yjqqaq"
this string contains plalindromes:"y","j","q","a","q","qq","qaq".
so we can choose "qq" and "qaq".

 

Input

The first line of input contains an positive integer  T(T<=10)  indicating the number of test cases.

For each test case:

First line contains these positive integer  N(1<=N<=100),K(1<=K<=100),L(L<=100) .
The next N line,each line contains a string only contains lowercase.Guarantee even length of string won't more than L.

 

Output

For each test,Output a line.If can output "True",else output "False".

 

Sample Input

   
   
   
   

    
    
    
    3
2 3 7
yjqqaq
claris
2 2 7
popoqqq
fwwf
1 3 3
aaa
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    False
True
True
   
   
   
   

刚好上个月学了回文树，这题可以把全部的串插进回文树里，然后统计每个有多少，然后就是多重背包问题了。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef unsigned long long LL;
const int maxn = 1e2 + 10;
int T, n, w, m, c[maxn], f[maxn][maxn];
char s[maxn];

struct PalindromicTree
{
	const static int maxn = 1e5 + 10;
	const static int size = 26;
	int next[maxn][size], last, sz, tot;
	int fail[maxn], len[maxn], cnt[maxn];
	char s[maxn];
	int Node(int length)
	{
		memset(next[sz], 0, sizeof(next[sz]));
		len[sz] = length; cnt[sz] = 0; return sz++;
	}
	void clear()
	{
		sz = last = fail[2] = fail[1] = 1;
		Node(-1);	Node(tot = 0);
	}
	void reset()
	{
		last = 1;	tot = 0;
	}
	int getfail(int x)
	{
		while (s[tot] != s[tot - len[x] - 1]) x = fail[x];
		return x;
	}
	void add(char pos)
	{
		int x = (s[++tot] = pos) - 'a', y = getfail(last);
		if (!(last = next[y][x]))
		{
			last = next[y][x] = Node(len[y] + 2);
			fail[last] = len[last] == 1 ? 2 : next[getfail(fail[y])][x];
		}
		++cnt[last];
	}
	void work()
	{
		memset(f, 0, sizeof(f));
		memset(c, 0, sizeof(c));
		for (int i = sz - 1; i > 2; i--)
		{
			cnt[fail[i]] += cnt[i];
			c[len[i]] += cnt[i];
		}
		f[0][0] = 1;
		for (int i = 1; i <= 100; i++)
		{
			for (int j = 1; c[i] > 0; j <<= 1)
			{
				int a = i*(j <= c[i] ? j : c[i]), b = j <= c[i] ? j : c[i];
				c[i] -= j;
				for (int k = w; k >= b; k--)
				{
					for (int kk = m; kk >= a; kk--)
					{
						f[k][kk] = f[k][kk] || f[k - b][kk - a];
					}
				}
			}
		}
		printf("%s\n", f[w][m] ? "True" : "False");
	}
}solve;

int main()
{
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d%d%d", &n, &w, &m);
		solve.clear();
		while (n--)
		{
			scanf("%s", s);		solve.reset();
			for (int i = 0; s[i]; i++) solve.add(s[i]);
		}
		solve.work();
	}
	return 0;
}