hdu3341 Resource Archiver AC自动机+DP
Lost's revenge
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 3128 Accepted Submission(s): 823
Problem Description
Lost and AekdyCoin are friends. They always play "number game"(A boring game based on number theory) together. We all know that AekdyCoin is the man called "nuclear weapon of FZU,descendant of Jingrun", because of his talent in the field of number theory. So Lost had never won the game. He was so ashamed and angry, but he didn't know how to improve his level of number theory.
One noon, when Lost was lying on the bed, the Spring Brother poster on the wall(Lost is a believer of Spring Brother) said hello to him! Spring Brother said, "I'm Spring Brother, and I saw AekdyCoin shames you again and again. I can't bear my believers were being bullied. Now, I give you a chance to rearrange your gene sequences to defeat AekdyCoin!".
It's soooo crazy and unbelievable to rearrange the gene sequences, but Lost has no choice. He knows some genes called "number theory gene" will affect one "level of number theory". And two of the same kind of gene in different position in the gene sequences will affect two "level of number theory", even though they overlap each other. There is nothing but revenge in his mind. So he needs you help to calculate the most "level of number theory" after rearrangement.
One noon, when Lost was lying on the bed, the Spring Brother poster on the wall(Lost is a believer of Spring Brother) said hello to him! Spring Brother said, "I'm Spring Brother, and I saw AekdyCoin shames you again and again. I can't bear my believers were being bullied. Now, I give you a chance to rearrange your gene sequences to defeat AekdyCoin!".
It's soooo crazy and unbelievable to rearrange the gene sequences, but Lost has no choice. He knows some genes called "number theory gene" will affect one "level of number theory". And two of the same kind of gene in different position in the gene sequences will affect two "level of number theory", even though they overlap each other. There is nothing but revenge in his mind. So he needs you help to calculate the most "level of number theory" after rearrangement.
Input
There are less than 30 testcases.
For each testcase, first line is number of "number theory gene" N(1<=N<=50). N=0 denotes the end of the input file.
Next N lines means the "number theory gene", and the length of every "number theory gene" is no more than 10.
The last line is Lost's gene sequences, its length is also less or equal 40.
All genes and gene sequences are only contains capital letter ACGT.
For each testcase, first line is number of "number theory gene" N(1<=N<=50). N=0 denotes the end of the input file.
Next N lines means the "number theory gene", and the length of every "number theory gene" is no more than 10.
The last line is Lost's gene sequences, its length is also less or equal 40.
All genes and gene sequences are only contains capital letter ACGT.
Output
For each testcase, output the case number(start with 1) and the most "level of number theory" with format like the sample output.
Sample Input
3 AC CG GT CGAT 1 AA AAA 0
Sample Output
Case 1: 3 Case 2: 2
给一个串，包含ACGT，重新排序后最多包含上面多少个串，可重叠。
既然可以重新排序，那么就只跟ACGT出现的次数有关了，可以想到开个5维数组进行DP，dp[41][41][41][41][50*10]，前4维分别表示每个字母个数，最后一维表示当前节点。但是这个会超内存的，注意那个串最长40，你每维都开40肯定是有浪费的。
类似于康托展开那种编码，用num[0],num[1],num[2],num[3]表示A,C,G,T的总个数，base表示基数，base[0]=(num[1]+1)*(num[2]+1)*(num[3]+1)，base[1]=(num[2]+1)*(num[3]+1)，base[2]=num3+1,base[3]=1，那么对于不超过总个数的任何ACGT的状态都有一一对应的编码，不会冲突。
这个问题解决了就容易了，转移方程不难。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<iostream>
#include<queue>
using namespace std;
typedef unsigned long long ULL;
const int MAXN=52;
const int MAXM=12;
const int MAXL=45;
const int MAXNODE=MAXN*MAXM;
const int LOGMAXN=50;
const int INF=0x3f3f3f3f;
const int SIGMA_SIZE=4;
const int MOD=20090717;
int T,N;
int dp[MAXNODE][14700];
int num[5],base[5];
struct AC{
int ch[MAXNODE][SIGMA_SIZE];
int val[MAXNODE];
int f[MAXNODE];
int sz;
void clear(){
memset(ch[0],0,sizeof(ch[0]));
val[0]=0;
sz=1;
}
int idx(char c){
switch(c){
case 'A':return 0;
case 'C':return 1;
case 'G':return 2;
case 'T':return 3;
}
}
void insert(char* s,int v){
int u=0;
for(int i=0;s[i];i++){
int c=idx(s[i]);
if(!ch[u][c]){
memset(ch[sz],0,sizeof(ch[sz]));
val[sz]=0;
ch[u][c]=sz++;
}
u=ch[u][c];
}
val[u]+=v;
}
void get_fail(){
queue<int> q;
f[0]=0;
for(int c=0;c<SIGMA_SIZE;c++){
int u=ch[0][c];
if(u){
f[u]=0;
q.push(u);
}
}
while(!q.empty()){
int r=q.front();
q.pop();
for(int c=0;c<SIGMA_SIZE;c++){
int u=ch[r][c];
if(!u){
ch[r][c]=ch[f[r]][c];
continue;
}
q.push(u);
f[u]=ch[f[r]][c];
val[u]+=val[f[u]];
}
}
}
}ac;
void DP(int len){
memset(dp,-1,sizeof(dp));
dp[0][0]=0;
for(int l=0;l<len;l++)
for(int i=0;i<=num[0]&&i<=l;i++)
for(int j=0;j<=num[1]&&j+i<=l;j++)
for(int k=0;k<=num[2]&&j+i+k<=l;k++){
int q=l-i-j-k;
if(q>num[3]) continue;
int n=i*base[0]+j*base[1]+k*base[2]+q*base[3];
for(int u=0;u<ac.sz;u++) if(dp[u][n]!=-1){
for(int c=0;c<SIGMA_SIZE;c++){
int p;
if(c==0) p=i+1;
else if(c==1) p=j+1;
else if(c==2) p=k+1;
else p=q+1;
if(p<=num[c]) dp[ac.ch[u][c]][n+base[c]]=max(dp[ac.ch[u][c]][n+base[c]],dp[u][n]+ac.val[ac.ch[u][c]]);
}
}
}
int ans=0;
int m=num[0]*base[0]+num[1]*base[1]+num[2]*base[2]+num[3]*base[3];
for(int u=0;u<ac.sz;u++) ans=max(ans,dp[u][m]);
printf("%d\n",ans);
}
char str[MAXM];
char P[MAXL];
int main(){
int cas=0;
while(scanf("%d",&N)!=EOF&&N){
ac.clear();
for(int i=0;i<N;i++){
scanf("%s",str);
ac.insert(str,1);
}
ac.get_fail();
memset(num,0,sizeof(num));
scanf("%s",P);
for(int i=0;P[i];i++) num[ac.idx(P[i])]++;
base[3]=1;
for(int i=2;i>=0;i--) base[i]=base[i+1]*(num[i+1]+1);
printf("Case %d: ",++cas);
DP(strlen(P));
}
return 0;
}