POJ 3691 DNA repair(AC自动机+DP)

题目链接

能AC还是很开心的...此题没有POJ2778那么难，那个题还需要矩阵乘法，两个题有点相似的。

做题之前，把2778代码重新看了一下，回忆一下当时做题的思路，回忆AC自动机是干嘛的...

状态表示dp[i][j]长度为i的以j串为结束的最小改变数目。AC自动机预处理一下，然后DP。

卡内存+不知道状态数，MLE+RE+WA了多次。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <cstdlib>
using namespace std;
#define N 2222222
#define INF 10000000
int trie[N][4];
int o[N];
int que[N];
int fail[N];
int dp[1001][1001];
int t;
void CL()
{
    memset(trie,-1,sizeof(trie));
    memset(o,0,sizeof(o));
    t = 1;
}
int judge(char s)
{
    switch(s)
    {
        case'A':return 0;
        case'C':return 1;
        case'G':return 2;
        case'T':return 3;
    }
    return 0;
}
void insert(char *str)
{
    int i,len,root;
    root = 0;
    len = strlen(str);
    for(i = 0;i < len;i ++)
    {
        if(trie[root][judge(str[i])] == -1)
        trie[root][judge(str[i])] = t ++;
        root = trie[root][judge(str[i])];
    }
    o[root] = 1;
}
void build_ac()
{
    int head,tail,front,i;
    head = tail = 0;
    for(i = 0;i < 4;i ++)
    {
        if(trie[0][i] != -1)
        {
            fail[trie[0][i]] = 0;
            que[tail++] = trie[0][i];
        }
        else
        {
            trie[0][i] = 0;
        }
    }
    while(head != tail)
    {
        front = que[head++];
        if(o[fail[front]])
        o[front] = 1;
        for(i = 0;i < 4;i ++)
        {
            if(trie[front][i] != -1)
            {
                que[tail++] = trie[front][i];
                fail[trie[front][i]] = trie[fail[front]][i];
            }
            else
            {
                trie[front][i] = trie[fail[front]][i];
            }
        }
    }
}
int main()
{
    int n,i,j,k,len,flag,cas = 1;
    char str[2001];
    while(scanf("%d",&n)!=EOF)
    {
        if(n == 0) break;
        CL();
        for(i = 1;i <= n;i ++)
        {
            scanf("%s",str);
            insert(str);
        }
        build_ac();
        scanf("%s",str);
        len = strlen(str);
        for(i = 0;i <= len;i ++)
        {
            for(j = 0;j <= t;j ++)
            dp[i][j] = INF;
        }
        dp[0][0] = 0;
        for(i = 0;i < len;i ++)
        {
            for(j = 0;j < t;j ++)
            {
                if(o[j]) continue;
                for(k = 0;k < 4;k ++)
                {
                    if(o[trie[j][k]]) continue;
                    flag = !(k == judge(str[i]));
                    dp[i+1][trie[j][k]] = min(dp[i+1][trie[j][k]],dp[i][j] + flag);
                }
            }
        }
        int ans = INF;
        for(i = 0;i < t;i ++)
        {
            ans = min(ans,dp[len][i]);
        }
        printf("Case %d: ",cas ++);
        if(ans == INF)
        printf("-1
");
        else
        printf("%d
",ans);
    }
    return 0;
}