Poj3691(AC自动机+DP(简单题))

//#pragma comment(linker, "/STACK:102400000")
#include<cstdlib>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<vector>
#define tree int o,int l,int r
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
#define lo o<<1
#define ro o<<1|1
#define pb push_back
#define mp make_pair
#define ULL unsigned long long
#define LL long long
#define inf 0x3fffffff//不要太大!溢出!
#define eps 1e-7
#define N 1009
#define M 4
#define mod 100000
using namespace std;
int m,n,T,t,x,y,u;
int ch[N][4];
int v[N];
int f[N],last[N],num;
char str[N];
int d[N][N];
void clear()//Trie树初始化
{
    memset(d,-1,sizeof(d));
    num=1;
    memset(ch[0],0,sizeof(ch[0]));
    memset(v,0,sizeof(v));
    memset(last,0,sizeof(last));
}
int idx(char c)
{
    switch(c)
    {
    case 'A':
        return 0;
    case 'C':
        return 1;
    case 'G':
        return 2;
    case 'T':
        return 3;
    }
    return 0;
}
void insert(char str[],int value)//建Trie树
{
    int len=strlen(str);
    int u=0;
    for (int i=0; i<len; ++i )
    {
        int c=idx(str[i]);
        if(!ch[u][c])//保存的是结点坐标
        {
            memset(ch[num],0,sizeof(ch[num]));
            ch[u][c]=num++;//
        }
        u=ch[u][c];
    }
    v[u]=value;
}
void getac()
{
    queue<int> q;//保存的节点下标
    f[0]=0;
    for (int c=0; c<M; ++c )
    {
        int u=ch[0][c];
        if(u)//不需要优化的else
        {
            q.push(u);
            f[u]=0;
            last[u]=v[u];//WA,可能有长度为1的串
        }
    }
    while(!q.empty())
    {
        int r=q.front();
        q.pop();
        for (int c=0; c<M; ++c )
        {
            int u=ch[r][c];
            if(u)
            {
                q.push(u);
                int s=f[r];
                f[u]=ch[s][c];
                last[u]=(v[u]||last[f[u]]);//改
            }
            else //重要优化
                ch[r][c]=ch[f[r]][c];
        }
    }
}
int dp(int u,int k)
{
    if(k==n)return 0;
    int &ans=d[u][k];
    if(ans!=-1)return ans;
    ans=inf;
    for(int i=0; i<4; i++)
    {
        int c=ch[u][i];
        if(last[c]==0)
        {
            ans=min(ans,dp(c,k+1)+(idx(str[k])!=i?1:0));
        }
    }
    return ans;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("ex.in","r",stdin);
#endif
    int ncase=0;
//    scanf("%d",&T);
//    while(T--)
    while(scanf("%d",&m)==1&&m)
    {
        clear();
        while(m--)
        {
            scanf("%s",str);
            insert(str,1);
        }
        getac();
        scanf("%s",str);
        n=strlen(str);
        printf("Case %d: %d\n",++ncase,dp(0,0)==inf?-1:dp(0,0));
    }
    return 0;
}

  

posted @ 2013-10-17 11:27  baoff  阅读(422)  评论(0编辑  收藏  举报