HDU - 3341 Lost's revenge(AC自己主动机+DP)

Description

Lost and AekdyCoin are friends. They always play "number game"(A boring game based on number theory) together. We all know that AekdyCoin is the man called "nuclear weapon of FZU,descendant of Jingrun", because of his talent in the
field of number theory. So Lost had never won the game. He was so ashamed and angry, but he didn't know how to improve his level of number theory. 



One noon, when Lost was lying on the bed, the Spring Brother poster on the wall(Lost is a believer of Spring Brother) said hello to him! Spring Brother said, "I'm Spring Brother, and I saw AekdyCoin shames you again and again. I can't bear my believers were
being bullied. Now, I give you a chance to rearrange your gene sequences to defeat AekdyCoin!". 



It's soooo crazy and unbelievable to rearrange the gene sequences, but Lost has no choice. He knows some genes called "number theory gene" will affect one "level of number theory". And two of the same kind of gene in different position in the gene sequences
will affect two "level of number theory", even though they overlap each other. There is nothing but revenge in his mind. So he needs you help to calculate the most "level of number theory" after rearrangement.

 

Input

There are less than 30 testcases. 

For each testcase, first line is number of "number theory gene" N(1<=N<=50). N=0 denotes the end of the input file. 

Next N lines means the "number theory gene", and the length of every "number theory gene" is no more than 10. 

The last line is Lost's gene sequences, its length is also less or equal 40. 

All genes and gene sequences are only contains capital letter ACGT. 

 

Output

For each testcase, output the case number(start with 1) and the most "level of number theory" with format like the sample output.

 

Sample Input

 3
AC
CG
GT
CGAT
1
AA
AAA
0 

 

Sample Output

 Case 1: 3
Case 2: 2 

 

题意：给你n个子串和一个母串，让你重排母串最多能得到多少个子串。

思路：如果ACGT的个数依次是num[0],num[1],num[2],num[3],将这四个数压缩成一个数，降低内存，然后动归dp[i][status]表示到自己主动机上i状态此时ACGT个数的状态时的最少

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;

struct Trie {
	int nxt[510][4], fail[510], end[510];
	int root, cnt;

	int newNode() {
		for (int i = 0; i < 4; i++)
			nxt[cnt][i] = -1;
		end[cnt++] = 0;
		return cnt - 1;
	}
	void init() {
		cnt = 0;
		root = newNode();
	}
	int getch(char ch) {
		if (ch == 'A') return 0;
		else if (ch == 'C') return 1;
		else if (ch == 'G') return 2;
		return 3;
	}
	void insert(char str[]) {
		int len = strlen(str);
		int now = root;
		for (int i = 0; i < len; i++) {
			if (nxt[now][getch(str[i])] == -1)
				nxt[now][getch(str[i])] = newNode();
			now = nxt[now][getch(str[i])];
		}
		end[now]++;
	}
	void build() {
		queue<int> q;
		fail[root] = root;
		for (int i = 0; i < 4; i++) {
			if (nxt[root][i] == -1)
				nxt[root][i] = root;
			else {
				fail[nxt[root][i]] = root;
				q.push(nxt[root][i]);
			}
		}

		while (!q.empty()) {
			int now = q.front();
			q.pop();
			end[now] += end[fail[now]]; //好像每次都是这里
			for (int i = 0; i < 4; i++) {
				if (nxt[now][i] == -1)
					nxt[now][i] = nxt[fail[now]][i];
				else {
					fail[nxt[now][i]] = nxt[fail[now]][i];
					q.push(nxt[now][i]);
				}
			}
		}
	}
	int dp[510][11*11*11*11+5];
	int bit[4], num[4];
	int solve(char str[]) {
		int len = strlen(str);
		memset(num, 0, sizeof(num));
		for (int i = 0; i < len; i++)
			num[getch(str[i])]++;
		bit[0] = (num[1]+1) * (num[2]+1) * (num[3]+1);
		bit[1] = (num[2]+1) * (num[3]+1);
		bit[2] = (num[3]+1);
		bit[3] = 1;
		memset(dp, -1, sizeof(dp));
		dp[root][0] = 0;
		for (int A = 0; A <= num[0]; A++)
			for (int B = 0; B <= num[1]; B++)
				for (int C = 0; C <= num[2]; C++)
					for (int D = 0; D <= num[3]; D++) {
						int s = A * bit[0] + B * bit[1] + C * bit[2] + D * bit[3];
						for (int i = 0; i < cnt; i++)
							if (dp[i][s] >= 0) {
								for (int k = 0; k < 4; k++) {
									if (k == 0 && A == num[0]) continue;
									if (k == 1 && B == num[1]) continue;
									if (k == 2 && C == num[2]) continue;
									if (k == 3 && D == num[3]) continue;
									dp[nxt[i][k]][s+bit[k]] = max(dp[nxt[i][k]][s+bit[k]], dp[i][s]+end[nxt[i][k]]);
								}
							}
					}

		int ans = 0;
		int status = num[0] * bit[0] + num[1] * bit[1] + num[2] * bit[2] + num[3] * bit[3];
		for (int i = 0; i < cnt; i++)
			ans = max(ans, dp[i][status]);
		return ans;
	}
} ac;
char str[50];

int main() {
	int n, cas = 1;
	while (scanf("%d", &n) != EOF && n) {
		ac.init();
		for (int i = 0; i < n; i++) {
			scanf("%s", str);
			ac.insert(str);
		}

		ac.build();
		scanf("%s", str);
		printf("Case %d: %d\n", cas++, ac.solve(str));
	}
	return 0;
}