DNA repair

#

Time Limit: 2000ms Memory limit: 32768K 有疑问？点这里^_^

题目描述

Biologists finally invent techniques of repairing DNA that contains segments causing kinds of inherited diseases. For the sake of simplicity, a DNA is represented as a string containing characters ‘A’, ‘G’ , ‘C’ and ‘T’. The repairing techniques are simply to change some characters to eliminate all segments causing diseases. For example, we can repair a DNA “AAGCAG” to “AGGCAC” to eliminate the initial causing disease segments “AAG”, “AGC” and “CAG” by changing two characters. Note that the repaired DNA can still contain only characters ‘A’, ‘G’, ‘C’ and ‘T’.

You are to help the biologists to repair a DNA by changing least number of characters.

输入

The input consists of multiple test cases. Each test case starts with a line containing one integers N (1 ≤ N ≤ 50), which is the number of DNA segments causing inherited diseases. The following N lines gives N non-empty strings of length not greater than 20 containing only characters in “AGCT”, which are the DNA segments causing inherited disease. The last line of the test case is a non-empty string of length not greater than 1000 containing only characters in “AGCT”, which is the DNA to be repaired.

The last test case is followed by a line containing one zeros.

输出

For each test case, print a line containing the test case number( beginning with 1) followed by the number of characters which need to be changed. If it’s impossible to repair the given DNA, print -1.

示例输入

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

示例输出

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

提示

hdoj2457 有链接提示的题目请先去链接处提交程序，AC后提交到SDUTOJ中，以便查询存档。

来源

2008 Asia Hefei Regional Contest Online by USTC

示例程序

include

#include<cstdio>    
#include<cstring>    
#include<cmath>    
#include<algorithm>    
#define N 100005    
#define MOD 100000    
#define inf 1<<29    
#define LL long long    
using namespace std;    
struct Trie{    
    Trie *next[4];    
    Trie *fail;    
    int kind,isword;    
};    
Trie *que[N],s[N];    
int idx;    
int id(char ch){    
    if(ch=='A') return 0;    
    else if(ch=='T') return 1;    
    else if(ch=='C') return 2;    
    return 3;    
}    
Trie *NewNode(){    
    Trie *tmp=&s[idx];    
    for(int i=0;i<4;i++) tmp->next[i]=NULL;    
    tmp->isword=0;    
    tmp->kind=idx++;    
    tmp->fail=NULL;    
    return tmp;    
}    
void Insert(Trie *root,char *s,int len){    
    Trie *p=root;    
    for(int i=0;i<len;i++){    
        if(p->next[id(s[i])]==NULL)    
            p->next[id(s[i])]=NewNode();    
        p=p->next[id(s[i])];    
    }    
    p->isword=1;    
}    
void Bulid_Fail(Trie *root){    
    int head=0,tail=0;    
    que[tail++]=root;    
    root->fail=NULL;    
    while(head<tail){    
        Trie *tmp=que[head++];    
        for(int i=0;i<4;i++){    
            if(tmp->next[i]){    
                if(tmp==root) tmp->next[i]->fail=root;    
                else{    
                    Trie *p=tmp->fail;    
                    while(p!=NULL){    
                        if(p->next[i]){    
                           tmp->next[i]->fail=p->next[i];    
                           break;    
                        }    
                        p=p->fail;    
                    }    
                    if(p==NULL) tmp->next[i]->fail=root;    
                }    
                if(tmp->next[i]->fail->isword) tmp->next[i]->isword=1;    
                que[tail++]=tmp->next[i];    
            }    
            else if(tmp==root) tmp->next[i]=root;    
            else tmp->next[i]=tmp->fail->next[i];    
        }    
    }    
}    
int dp[1005][2005];    
int slove(char *str,int len){    
    for(int i=0;i<=len;i++) for(int j=0;j<idx;j++) dp[i][j]=inf;    
    dp[0][0]=0;    
    for(int i=1;i<=len;i++){    
        for(int j=0;j<idx;j++){    
            if(s[j].isword) continue;    
            if(dp[i-1][j]==inf) continue;    
            for(int k=0;k<4;k++){    
                int r=s[j].next[k]->kind;    
                if(s[r].isword) continue;    
                dp[i][r]=min(dp[i][r],dp[i-1][j]+(id(str[i-1])!=k));    
            }    
        }    
    }    
    int ans=inf;    
    for(int i=0;i<idx;i++) ans=min(ans,dp[len][i]);    
    return ans==inf?-1:ans;    
}    
char str[1005];    
int main(){    
    int n,cas=0;    
    while(scanf("%d",&n)!=EOF&&n){    
        idx=0;    
        Trie *root=NewNode();    
        for(int i=0;i<n;i++){    
            scanf("%s",str);    
            Insert(root,str,strlen(str));    
        }    
        Bulid_Fail(root);    
        scanf("%s",str);    
        printf("Case %d: %d\n",++cas,slove(str,strlen(str)));    
    }    
    return 0;    
}