2014
01-05

# Ring

For the hope of a forever love, Steven is planning to send a ring to Jane with a romantic string engraved on. The string’s length should not exceed N. The careful Steven knows Jane so deeply that he knows her favorite words, such as "love", "forever". Also, he knows the value of each word. The higher value a word has the more joy Jane will get when see it.
The weight of a word is defined as its appeared times in the romantic string multiply by its value, while the weight of the romantic string is defined as the sum of all words’ weight. You should output the string making its weight maximal.

The input consists of several test cases. The first line of input consists of an integer T, indicating the number of test cases. Each test case starts with a line consisting of two integers: N, M, indicating the string’s length and the number of Jane’s favorite words. Each of the following M lines consists of a favorite word Si. The last line of each test case consists of M integers, while the i-th number indicates the value of Si.
Technical Specification

1. T ≤ 15
2. 0 < N ≤ 50, 0 < M ≤ 100.
3. The length of each word is less than 11 and bigger than 0.
4. 1 ≤ Hi ≤ 100.
5. All the words in the input are different.
6. All the words just consist of ‘a’ – ‘z’.

The input consists of several test cases. The first line of input consists of an integer T, indicating the number of test cases. Each test case starts with a line consisting of two integers: N, M, indicating the string’s length and the number of Jane’s favorite words. Each of the following M lines consists of a favorite word Si. The last line of each test case consists of M integers, while the i-th number indicates the value of Si.
Technical Specification

1. T ≤ 15
2. 0 < N ≤ 50, 0 < M ≤ 100.
3. The length of each word is less than 11 and bigger than 0.
4. 1 ≤ Hi ≤ 100.
5. All the words in the input are different.
6. All the words just consist of ‘a’ – ‘z’.

2
7 2
love
ever
5 5
5 1
ab
5

lovever
abab

Hint
Sample 1: weight(love) = 5, weight(ever) = 5, so weight(lovever) = 5 + 5 = 10
Sample 2: weight(ab) = 2 * 5 = 10, so weight(abab) = 10


#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<queue>
#include<algorithm>
#include<vector>
#include<stack>
#include<list>
#include<iostream>
#include<map>
using namespace std;
#define inf 0x3f3f3f3f
#define Max 110
int max(int a,int b)
{
return a>b?a:b;
}
int min(int a,int b)
{
return a<b?a:b;
}
int q[12*110],cnt;
char str[110][12];
int dp[60][1100];
int pre[60][1100];
char  c[1100];
struct node
{
int cnt,fail,al;
int next[26];
void init()
{
cnt=fail=0;
memset(next,0,sizeof(next));
}
}tri[12*110];
int t,n,m;
inline bool cmp(int i,int x,int y)
{
for(int k=i;k>=1;k--)
{
if(c[x]!=c[y])
return c[x]<c[y];
x=pre[k][x];
y=pre[k][y];
}
return false;
}
void insert(char *s,int val)
{
int i,p,x;
p=0;
for(i=0;s[i];i++)
{
x=s[i]-'a';
if(!tri[p].next[x])
{
tri[++cnt].init();
tri[cnt].al=s[i];
tri[p].next[x]=cnt;
c[cnt]=s[i];
}
p=tri[p].next[x];
}
tri[p].cnt=val;
}
void bfs()
{
int i,p=0,suf,head=0,tail=0;
for(i=0;i<26;i++)
{
if(tri[0].next[i])
{
q[tail++]=tri[0].next[i];
tri[q[tail-1]].fail=0;
}
}
while(head<tail)
{
p=q[head++];suf=tri[p].fail;
tri[p].cnt+=tri[suf].cnt;
for(i=0;i<26;i++)
{
if(tri[p].next[i])
{
q[tail++]=tri[p].next[i];
tri[q[tail-1]].fail=tri[suf].next[i];
}
else
tri[p].next[i]=tri[suf].next[i];
}
}
}
void solve()
{
int ans=0,len=0,pos=0,i,j,k,tmp;
for(i=0;i<=n;i++)
for(j=0;j<=cnt;j++)
dp[i][j]=-1;
dp[0][0]=0;
for(i=0;i<=n;i++)
{
for(j=0;j<=cnt;j++)
{
if((i&&(j==0))||dp[i][j]==-1)
continue;
if(i&&(ans<dp[i][j]))
{
ans=dp[i][j];
len=i;
pos=j;
}
else if(i&&(ans==dp[i][j]))
{
if(len==i&&cmp(len,j,pos))
pos=j;
}
if(i==n)
continue;
for(k=0;k<26;k++)
{
tmp=tri[j].next[k];
if(tmp==0)
continue;
if(dp[i+1][tmp]<dp[i][j]+tri[tmp].cnt)
{
dp[i+1][tmp]=dp[i][j]+tri[tmp].cnt;
pre[i+1][tmp]=j;
}
else if(dp[i+1][tmp]==dp[i][j]+tri[tmp].cnt)
{
if(cmp(i,j,pre[i+1][tmp]))
pre[i+1][tmp]=j;
}
}
}
}
if(ans==0)
{
puts("");
return ;
}
for(i=len;i>0;i--)
{
printf("%c",c[pos]);
pos=pre[i][pos];
}
puts("");
}
int main()
{
//freopen("D:/ring.txt","w",stdout);
int v,i,len;
char s[20];
scanf("%d",&t);
while(t--)
{
cnt=0;tri[0].init();
scanf("%d%d",&n,&m);
for(i=0;i<m;i++)
scanf("%s",str[i]);
for(i=0;i<m;i++)
{
scanf("%d",&v);
len=strlen(str[i]);
for(int j=len-1;j>=0;j--)
s[len-1-j]=str[i][j];
s[len]=0;
insert(s,v);
}
bfs();
solve();
}
}