HDU - 3341 Lost's revenge(AC自己主动机+DP)

 3
AC
CG
GT
CGAT
1
AA
AAA
0 

 Case 1: 3
Case 2: 2 

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;
}