hdu3341Lost's revenge (AC自动机+变进制dp)
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3657 Accepted Submission(s): 987
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.
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.
AC
CG
GT
CGAT
1
AA
AAA
0
Case 2: 2
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define pi acos(-1.0)
#define maxnode 500000
int t0,t1,t2,t3;
char s[14],str[50];
int cas=0;
int dp[505][15005];
struct trie{
int sz,root,val[maxnode],next[maxnode][4],fail[maxnode];
int q[1111111];
void init(){
int i;
sz=root=0;
val[0]=0;
for(i=0;i<4;i++){
next[root][i]=-1;
}
}
int idx(char c){
if(c=='A')return 0;
if(c=='C')return 1;
if(c=='T')return 2;
if(c=='G')return 3;
}
void charu(char *s){
int i,j,u=0;
int len=strlen(s);
for(i=0;i<len;i++){
int c=idx(s[i]);
if(next[u][c]==-1){
sz++;
val[sz]=0;
next[u][c]=sz;
u=next[u][c];
for(j=0;j<4;j++){
next[u][j]=-1;
}
}
else{
u=next[u][c];
}
}
val[u]++;
}
void build(){
int i,j;
int front,rear;
front=1;rear=0;
for(i=0;i<4;i++){
if(next[root][i]==-1 ){
next[root][i]=root;
}
else{
fail[next[root][i] ]=root;
rear++;
q[rear]=next[root][i];
}
}
while(front<=rear){
int x=q[front];
val[x]+=val[fail[x] ];
front++;
for(i=0;i<4;i++){
if(next[x][i]==-1){
next[x][i]=next[fail[x] ][i];
}
else{
fail[next[x][i] ]=next[fail[x] ][i];
rear++;
q[rear]=next[x][i];
}
}
}
}
void chazhao(){
int i,j;
int bit[4],num0,num1,num2,num3,state,state1;
for(i=0;i<=sz;i++){
for(j=0;j<=15000;j++){
dp[i][j]=-1;
}
}
dp[0][0]=0;
bit[0]=(t1+1)*(t2+1)*(t3+1);
bit[1]=(t2+1)*(t3+1);
bit[2]=t3+1;
bit[3]=1;
for(num0=0;num0<=t0;num0++){
for(num1=0;num1<=t1;num1++){
for(num2=0;num2<=t2;num2++){
for(num3=0;num3<=t3;num3++){
state=bit[0]*num0+bit[1]*num1+bit[2]*num2+bit[3]*num3;
for(i=0;i<=sz;i++){
if(dp[i][state]==-1)continue;
if(num0<t0){
state1=state+bit[0];
dp[next[i][0] ][state1]=max(dp[next[i][0] ][state1],dp[i][state]+val[next[i][0] ] );
}
if(num1<t1){
state1=state+bit[1];
dp[next[i][1] ][state1]=max(dp[next[i][1] ][state1],dp[i][state]+val[next[i][1] ] );
}
if(num2<t2){
state1=state+bit[2];
dp[next[i][2] ][state1]=max(dp[next[i][2] ][state1],dp[i][state]+val[next[i][2] ] );
}
if(num3<t3){
state1=state+bit[3];
dp[next[i][3] ][state1]=max(dp[next[i][3] ][state1],dp[i][state]+val[next[i][3] ] );
}
}
}
}
}
}
int maxx=0;
state=t0*bit[0]+t1*bit[1]+t2*bit[2]+t3*bit[3];
for(i=0;i<=sz;i++){
maxx=max(maxx,dp[i][state]);
}
printf("Case %d: %d\n",++cas,maxx);
}
}ac;
int main()
{
int n,m,i,j;
while(scanf("%d",&n)!=EOF && n!=0){
ac.init();
for(i=1;i<=n;i++){
scanf("%s",s);
ac.charu(s);
}
ac.build();
scanf("%s",str);
t0=t1=t2=t3=0;
int len=strlen(str);
for(i=0;i<len;i++){
if(str[i]=='A')t0++;
else if(str[i]=='C')t1++;
else if(str[i]=='T')t2++;
else t3++;
}
ac.chazhao();
}
return 0;
}
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