Dr. X is a biologist, who likes rabbits very much and can do everything for them. 2012 is coming, and Dr. X wants to take some rabbits to Noah’s Ark, or there are no rabbits any more.
A rabbit’s genes can be expressed as a string whose length is l (1 ≤ l ≤ 100) containing only ‘A’, ‘G’, ‘T’, ‘C’. There is no doubt that Dr. X had a in-depth research on the rabbits’ genes. He found that if a rabbit gene contained a particular gene segment, we could consider it as a good rabbit, or sometimes a bad rabbit. And we use a value W to measure this index.
We can make a example, if a rabbit has gene segment “ATG”, its W would plus 4; and if has gene segment “TGC”, its W plus -3. So if a rabbit’s gene string is “ATGC”, its W is 1 due to ATGC contains both “ATG”(+4) and “TGC”(-3). And if another rabbit’s gene string is “ATGATG”, its W is 4 due to one gene segment can be calculate only once.
Because there are enough rabbits on Earth before 2012, so we can assume we can get any genes with different structure. Now Dr. X want to find a rabbit whose gene has highest W value. There are so many different genes with length l, and Dr. X is not good at programming, can you help him to figure out the W value of the best rabbit.
Input
There are multiple test cases. For each case the first line is two integers n (1 ≤ n ≤ 10)，l (1 ≤ l ≤ 100), indicating the number of the particular gene segment and the length of rabbits’ genes.
The next n lines each line contains a string DNAi and an integer wi (|wi| ≤ 100), indicating this gene segment and the value it can contribute to a rabbit’s W.
Output
For each test case, output an integer indicating the W value of the best rabbit. If we found this value is negative, you should output “No Rabbit after 2012!”.
Sample Input
2 4
ATG 4
TGC -3
1 6
TGC 4
4 1
A -1
T -2
G -3
C -4
Sample Output
4
4
No Rabbit after 2012!
Hint
case 1：we can find a rabbit whose gene string is ATGG(4), or ATGA(4) etc.
case 2：we can find a rabbit whose gene string is TGCTGC(4), or TGCCCC(4) etc.
case 3：any gene string whose length is 1 has a negative W.
Author: HONG, Qize
Contest: The 2011 ACM-ICPC Asia Dalian Regional Contest
还是很简单的一题dp, dp[i][j][k]表示长度为i,在节点j,状态为k的可行性,然后建立自动机转移就行,数组开不下,所以要用滚动数组,但是记得用滚动数组时,上次用的不能干扰到这次用的
/*************************************************************************
> File Name: ZOJ3545.cpp
> Author: ALex
> Mail: [email protected]
> Created Time: 2015年04月20日 星期一 15时42分47秒
************************************************************************/
#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>
using namespace std;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
const int MAX_NODE = 1010;
const int CHILD_NUM = 4;
bool dp[2][MAX_NODE][1100];
int mp[15];
struct AC_Automation
{
int next[MAX_NODE][CHILD_NUM];
int fail[MAX_NODE];
int end[MAX_NODE];
int root, L;
int newnode()
{
for (int i = 0; i < CHILD_NUM; ++i)
{
next[L][i] = -1;
}
end[L++] = 0;
return L - 1;
}
void init ()
{
L = 0;
root = newnode();
}
int ID(char c)
{
if (c == ‘A‘)
{
return 0;
}
if (c == ‘G‘)
{
return 1;
}
if (c == ‘T‘)
{
return 2;
}
return 3;
}
void Build_Trie (char buf[], int id)
{
int now = root;
int len = strlen (buf);
for (int i = 0; i < len; ++i)
{
if (next[now][ID(buf[i])] == -1)
{
next[now][ID(buf[i])] = newnode();
}
now = next[now][ID(buf[i])];
}
end[now] |= (1 << id);
}
void Build_AC ()
{
queue <int> qu;
fail[root] = root;
for (int i = 0; i < CHILD_NUM; ++i)
{
if (next[root][i] == -1)
{
next[root][i] = root;
}
else
{
fail[next[root][i]] = root;
qu.push (next[root][i]);
}
}
while (!qu.empty())
{
int now = qu.front();
qu.pop();
end[now] |= end[fail[now]];
for (int i = 0; i < CHILD_NUM; ++i)
{
if (next[now][i] == -1)
{
next[now][i] = next[fail[now]][i];
}
else
{
fail[next[now][i]] = next[fail[now]][i];
qu.push(next[now][i]);
}
}
}
}
void solve(int l, int n)
{
memset(dp, 0, sizeof(dp));
dp[0][0][0] = 1;
int now = 1;
for (int i = 0; i < l; ++i)
{
memset(dp[now], 0, sizeof(dp[now]));
for (int j = 0; j < L; ++j)
{
for (int k = 0; k < (1 << n); ++k)
{
for (int q = 0; q < 4; ++q)
{
int nxt = next[j][q];
dp[now][nxt][k | end[nxt]] = (dp[now][nxt][k | end[nxt]] || dp[now ^ 1][j][k]);
}
}
}
now ^= 1;
}
int ans = -inf;
for (int i = 0; i < L; ++i)
{
for (int j = 0; j < (1 << n); ++j)
{
if (dp[now ^ 1][i][j])
{
int w = 0;
for (int k = 0; k < n; ++k)
{
if (j & (1 << k))
{
w += mp[k];
}
}
ans = max(ans, w);
}
}
}
if (ans < 0)
{
printf("No Rabbit after 2012!\n");
}
else
{
printf("%d\n", ans);
}
}
}AC;
char str[110];
int main()
{
int n, l, w;
while (~scanf("%d%d", &n, &l))
{
AC.init();
for (int i = 0; i < n; ++i)
{
scanf("%s%d", str, &w);
AC.Build_Trie(str, i);
mp[i] = w;
}
AC.Build_AC();
AC.solve(l, n);
}
return 0;
}