[poj3691] DNA repair

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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.

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输入格式

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.

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输出格式

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.

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样例数据

样例输入

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

样例输出

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

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题目分析

和poj1625差不多，就不细讲了。
建立AC自动机方法一样，动规思路一样，不需要高精度，不需要调试，1A棒棒的

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源代码

#include<algorithm>#include<iostream>#include<iomanip>#include<cstring>#include<cstdlib>#include<vector>#include<cstdio>#include<cmath>#include<queue>using namespace std;inline const int Get_Int() {    int num=0,bj=1;    char x=getchar();    while(x<'0'||x>'9') {        if(x=='-')bj=-1;        x=getchar();    }    while(x>='0'&&x<='9') {        num=num*10+x-'0';        x=getchar();    }    return num*bj;}const int maxn=1005;struct Tree {    int child[15],fail,flag; //fail失败指针    void clear() {        memset(child,0,sizeof(child));        fail=0;        flag=0;    }};int root=1,n,Map[1005];struct Aho_Corasick_Automaton { //AC自动机    int cnt;    Tree tree[maxn];    void init() {        cnt=1;        memset(tree,0,sizeof(tree));    }    void insert(string s) {        int now=root,len=s.length();        for(int i=0; i<len; i++) {            int j=Map[s[i]];            if(!tree[now].child[j]) {                tree[++cnt].clear();                tree[now].child[j]=cnt;            }            now=tree[now].child[j];        }        tree[now].flag=1;    }    void buildfail() { //Bfs构造Fail指针        queue<int>Q;        Q.push(root);         while(!Q.empty()) {            int Now=Q.front();            Q.pop();            for(int i=0; i<4; i++) {                int Next=tree[Now].child[i];                if(Next==0) { //儿子不存在                    if(tree[tree[Now].fail].child[i])tree[Now].child[i]=tree[tree[Now].fail].child[i];                    else tree[Now].child[i]=root;                    continue;                }                Q.push(Next);                int fatherfail=tree[Now].fail; //父亲的失败指针                while(fatherfail&&!tree[fatherfail].child[i])fatherfail=tree[fatherfail].fail; //寻找可退回点                if(fatherfail)tree[Next].fail=tree[fatherfail].child[i]; //如果存在满足条件的点则设置失败指针                else tree[Next].fail=root; //否则指回root                tree[Next].flag|=tree[tree[Next].fail].flag;            }        }    }};Aho_Corasick_Automaton ac;long long f[1005][1005],ans=0x7fffffff/2,tot=0;string DNA;int main() {    ios::sync_with_stdio(false);    Map['A']=0;    Map['G']=1;    Map['C']=2;    Map['T']=3;    while(true) {        cin>>n;        if(n==0)break;        ans=0x7fffffff/2;        ac.init();        for(int i=1; i<=n; i++) {            string s;            cin>>s;            ac.insert(s);        }        cin>>DNA;        int len=DNA.length();        ac.buildfail();        for(int i=0; i<=len; i++)            for(int j=1; j<=ac.cnt; j++)                f[i][j]=0x7fffffff/2;        f[0][1]=0;        for(int i=1; i<=len; i++)            for(int j=1; j<=ac.cnt; j++) {                if(ac.tree[j].flag)continue;                for(int k=0; k<4; k++) {                    if(ac.tree[ac.tree[j].child[k]].flag)continue;                    f[i][ac.tree[j].child[k]]=min(f[i][ac.tree[j].child[k]],f[i-1][j]+(Map[DNA[i-1]]!=k));                }            }        printf("Case %d: ",++tot);        for(int i=1; i<=ac.cnt; i++)ans=min(ans,f[len][i]);        if(ans!=0x7fffffff/2)printf("%lld\n",ans);        else puts("-1");    }    return 0;}

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