POJ 3691 DNA repair (AC自動機 + dp)

Description

Input

Output

Sample Input

2
AAA
AAG
AAAG    
2
A
TG
TGAATG
4
A
G
C
T
AGT
0

Sample Output

Case 1: 1
Case 2: 4
Case 3: -1

Source

2008 Asia Hefei Regional Contest Online by USTC

題目連結：http://poj.org/problem?id=3691

題目大意：給一些壞的字串，要求主串中不出現壞的字串，問最少要改變幾個主串中的字元

題目分析：將問題變為在 AC機上走(不走過危險結點)，使得走過的路徑得到的串與目標串匹配最多，建AC機的時候如果一個點失敗指標指的點是壞點，那它也是壞點，相當於這個壞字串是另一個的子串，這裡還需要用Trie圖優化，方便dp時候狀態的轉移，dp[i][j]表示主串匹配了前i個字元且自動機在j結點時最少要改變的字元數，注意dp的時候要避開壞點

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
int const MAX = 1005;
int const INF = 1 << 30;
char s[50], t[MAX];

struct AC_DP
{
	int tot, root, fail[MAX], next[MAX][4];
	int dp[MAX][MAX];
	bool bad[MAX];

	inline int chg(char ch)
	{
		if(ch == 'A')
			return 0;
		if(ch == 'T')
			return 1;
		if(ch == 'C')
			return 2;
		return 3;
	}

	inline int Newnode()
	{
		for(int i = 0; i < 4; i++)
			next[tot][i] = -1;
		fail[tot] = 0;
		bad[tot] = false;
		return tot ++;
	}

	inline void Init()
	{
		tot = 0;
		root = Newnode();
	}

	inline void Insert(char *s)
	{
		int len = strlen(s);
		int p = root;
		for(int i = 0; i < len; i++)
		{
			int idx = chg(s[i]);
			if(next[p][idx] == -1)
				next[p][idx] = Newnode();
			p = next[p][idx];
		}
		bad[p] = true;
	}

	inline void Build()
	{
		queue <int> q;
		q.push(root);
		while(!q.empty())
		{
			int p = q.front();
			q.pop();
			for(int i = 0; i < 4; i++)
			{
				if(next[p][i] == -1)
				{
					if(p == root)
						next[p][i] = root;
					else
						next[p][i] = next[fail[p]][i];
				}
				else
				{
					if(p == root)
						fail[next[p][i]] = root;
					else
					{
						fail[next[p][i]] = next[fail[p]][i];
						bad[p] |= bad[fail[p]];
					}
					q.push(next[p][i]);
				}
			}
		}
	}

	inline int DP()
	{
		int tlen = strlen(t);
		int ans = INF;
		for(int i = 0; i <= tlen; i++)
			for(int j = 0; j < tot; j++)
				dp[i][j] = INF;
		dp[0][0] = 0;
		for(int i = 1; i <= tlen; i++)
			for(int j = 0; j < tot; j++)
				if(dp[i - 1][j] < INF)
					for(int k = 0; k < 4; k++)
						if(!bad[next[j][k]])
							dp[i][next[j][k]] = min(dp[i][next[j][k]], dp[i - 1][j] + (chg(t[i - 1]) != k));
		for(int j = 0; j < tot; j++)
			if(!bad[j])
				ans = min(ans, dp[tlen][j]);
		if(ans == INF)
			return -1;
		return ans;
	}	
}ac;

int main()
{
	int n, ca = 1;
	while(scanf("%d", &n) != EOF && n)
	{
		ac.Init();
		for(int i = 0; i < n; i++)
		{
			scanf("%s", s);
			ac.Insert(s);
		}
		scanf("%s", t);	
		ac.Build();
		printf("Case %d: %d\n", ca ++, ac.DP());
	}
}