Rescue the Rabbit
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2036 Accepted Submission(s): 591
Problem Description
Dr. X is a biologist, who likes rabbits very much and can do everything for them. 2012 is coming, and Dr. X wants to take some rabbits to Noah's Ark, or there are no rabbits any more.
A rabbit's genes can be expressed as a string whose length is l (1 ≤ l ≤ 100) containing only 'A', 'G', 'T', 'C'. There is no doubt that Dr. X had a in-depth research on the rabbits' genes. He found that if a rabbit gene contained a particular gene segment, we could consider it as a good rabbit, or sometimes a bad rabbit. And we use a value W to measure this index.
We can make a example, if a rabbit has gene segment "ATG", its W would plus 4; and if has gene segment "TGC", its W plus -3. So if a rabbit's gene string is "ATGC", its W is 1 due to ATGC contains both "ATG"(+4) and "TGC"(-3). And if another rabbit's gene string is "ATGATG", its W is 4 due to one gene segment can be calculate only once.
Because there are enough rabbits on Earth before 2012, so we can assume we can get any genes with different structure. Now Dr. X want to find a rabbit whose gene has highest W value. There are so many different genes with length l, and Dr. X is not good at programming, can you help him to figure out the W value of the best rabbit.
A rabbit's genes can be expressed as a string whose length is l (1 ≤ l ≤ 100) containing only 'A', 'G', 'T', 'C'. There is no doubt that Dr. X had a in-depth research on the rabbits' genes. He found that if a rabbit gene contained a particular gene segment, we could consider it as a good rabbit, or sometimes a bad rabbit. And we use a value W to measure this index.
We can make a example, if a rabbit has gene segment "ATG", its W would plus 4; and if has gene segment "TGC", its W plus -3. So if a rabbit's gene string is "ATGC", its W is 1 due to ATGC contains both "ATG"(+4) and "TGC"(-3). And if another rabbit's gene string is "ATGATG", its W is 4 due to one gene segment can be calculate only once.
Because there are enough rabbits on Earth before 2012, so we can assume we can get any genes with different structure. Now Dr. X want to find a rabbit whose gene has highest W value. There are so many different genes with length l, and Dr. X is not good at programming, can you help him to figure out the W value of the best rabbit.
Input
There are multiple test cases. For each case the first line is two integers n (1 ≤ n ≤ 10)，l (1 ≤ l ≤ 100), indicating the number of the particular gene segment and the length of rabbits' genes.
The next n lines each line contains a string DNAi and an integer wi (|wi| ≤ 100), indicating this gene segment and the value it can contribute to a rabbit's W.
The next n lines each line contains a string DNAi and an integer wi (|wi| ≤ 100), indicating this gene segment and the value it can contribute to a rabbit's W.
Output
For each test case, output an integer indicating the W value of the best rabbit. If we found this value is negative, you should output "No Rabbit after 2012!".
Sample Input
2 4
ATG 4
TGC -3
1 6
TGC 4
4 1
A -1
T -2
G -3
C -4
Sample Output
4
4
No Rabbit after 2012!
题目大意：给n个字符串，每个都有权值，其他的字符串没有权值。构造一个长度为m的权值最大的字符串，输出最大权值。每个带权字符串只能计算一次。
题目分析：容易想到在AC自动机上动态规划。内存要求比较高，用下滚动数组。
代码一：
# include<iostream>
# include<cstdio>
# include<queue>
# include<cstring>
# include<algorithm>
using namespace std;
const short N=1005;
const short INF=0x7fff;
short ch[N][4];
short sz;
short val[N];
void init()
{
sz=0;
memset(ch,-1,sizeof(ch));
memset(val,0,sizeof(val));
}
short idx(char c)
{
if(c=='A') return 0;
else if(c=='T') return 1;
else if(c=='C') return 2;
else if(c=='G') return 3;
}
void insert(char *s,short x)
{
short r=0;
short n=strlen(s);
for(short i=0;i<n;++i){
short c=idx(s[i]);
if(ch[r][c]==-1) ch[r][c]=++sz;
r=ch[r][c];
}
val[r]=x;
}
void getFail()
{
short *fail=new short[sz+1];
queue<short>q;
fail[0]=0;
for(short i=0;i<4;++i){
if(ch[0][i]==-1)
ch[0][i]=0;
else{
fail[ch[0][i]]=0;
q.push(ch[0][i]);
}
}
while(!q.empty())
{
short u=q.front();
q.pop();
val[u]|=val[fail[u]];
for(short i=0;i<4;++i){
if(ch[u][i]==-1)
ch[u][i]=ch[fail[u]][i];
else{
fail[ch[u][i]]=ch[fail[u]][i];
q.push(ch[u][i]);
}
}
}
delete []fail;
}
char s[6];
void solve(short n,short m,short *w)
{
short*** dp=new short**[2];
for(short i=0;i<2;++i){
dp[i]=new short*[sz+1];
for(short j=0;j<=sz;++j){
dp[i][j]=new short[1<<n];
}
}
for(short i=0;i<=sz;++i) for(short j=0;j<(1<<n);++j)
dp[0][i][j]=-INF;
dp[0][0][0]=0;
short flag=1;
for(short i=0;i<m;++i,flag^=1){
for(short j=0;j<=sz;++j)
for(short k=0;k<(1<<n);++k)
dp[flag][j][k]=-INF;
for(short j=0;j<=sz;++j){
for(short k=0;k<(1<<n);++k){
if(dp[flag^1][j][k]==-INF)
continue;
for(short c=0;c<4;++c){
short v=ch[j][c];
short tempw=0;
for(short l=0;l<n;++l){
if(!(k&(1<<l))&&(val[v]&(1<<l)))
tempw+=w[l];
}
short &nxt=dp[flag][v][k|val[v]];
if(nxt<dp[flag^1][j][k]+tempw)
nxt=dp[flag^1][j][k]+tempw;
}
}
}
}
short ans=-INF;
for(short i=0;i<=sz;++i)
for(short j=0;j<(1<<n);++j)
ans=max(ans,dp[flag^1][i][j]);
if(ans>=0)
cout<<ans<<endl;
else
printf("No Rabbit after 2012!
");
for(short i=0;i<2;++i){
for(short j=0;j<=sz;++j){
delete []dp[i][j];
}
delete []dp[i];
}
delete dp;
}
int main()
{
short n,m;
while(cin>>n>>m)
{
init();
short *w=new short[n];
for(short i=0;i<n;++i){
scanf("%s%d",s,w+i);
insert(s,1<<i);
}
getFail();
solve(n,m,w);
delete []w;
}
return 0;
}
代码二：
# include<iostream>
# include<cstdio>
# include<queue>
# include<cstring>
# include<algorithm>
using namespace std;
const int N=1005;
const int INF=0x7fff;
int ch[N][4];
int sz;
int val[N];
void init()
{
sz=0;
memset(ch,-1,sizeof(ch));
memset(val,0,sizeof(val));
}
int idx(char c)
{
if(c=='A') return 0;
else if(c=='T') return 1;
else if(c=='C') return 2;
else if(c=='G') return 3;
}
void insert(char *s,int x)
{
int r=0;
int n=strlen(s);
for(int i=0;i<n;++i){
int c=idx(s[i]);
if(ch[r][c]==-1) ch[r][c]=++sz;
r=ch[r][c];
}
val[r]=x;
}
void getFail()
{
int *fail=new int[sz+1];
queue<int>q;
fail[0]=0;
for(int i=0;i<4;++i){
if(ch[0][i]==-1)
ch[0][i]=0;
else{
fail[ch[0][i]]=0;
q.push(ch[0][i]);
}
}
while(!q.empty())
{
int u=q.front();
q.pop();
val[u]|=val[fail[u]];
for(int i=0;i<4;++i){
if(ch[u][i]==-1)
ch[u][i]=ch[fail[u]][i];
else{
fail[ch[u][i]]=ch[fail[u]][i];
q.push(ch[u][i]);
}
}
}
delete []fail;
}
char s[6];
int w[10];
int dp[2][N+1][1<<10];
int get(int s,int n)
{
int res=0;
for(int i=0;i<n;++i)
if(s&(1<<i)) res+=w[i];
return res;
}
void solve(int n,int m)
{
memset(dp,0,sizeof(dp));
dp[0][0][0]=1;
int flag=1;
for(int i=0;i<m;++i,flag^=1){
memset(dp[flag],0,sizeof(dp[flag]));
for(int j=0;j<=sz;++j){
for(int k=0;k<(1<<n);++k) if(dp[flag^1][j][k]){
for(int c=0;c<4;++c)
dp[flag][ch[j][c]][k|val[ch[j][c]]]=1;
}
}
}
int ans=-INF;
for(int s=0;s<(1<<n);++s){
int temp=get(s,n);
for(int i=0;i<=sz;++i){
if(dp[flag^1][i][s]&&temp>ans){
ans=temp;
break;
}
}
}
if(ans>=0)
cout<<ans<<endl;
else
printf("No Rabbit after 2012!
");
}
int main()
{
int n,m;
while(cin>>n>>m)
{
init();
for(int i=0;i<n;++i){
scanf("%s%d",s,w+i);
insert(s,1<<i);
}
getFail();
solve(n,m);
}
return 0;
}