hdu4057Rescue the Rabbit(ac自动机+dp)

链接

当时是因为没有做出来这道题才开了自动机的专题,现在看看还是比较简单的。

因为每个病毒串只算一次,只有10个病毒串,可以状压一下哪些状态是可以达到的,最后取一个最大值。

 #include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
#include<string>
using namespace std;
#define N 1005
#define LL long long
#define INF 0xfffffff
const double eps = 1e-;
const double pi = acos(-1.0);
const double inf = ~0u>>;
const int child_num = ;
char vir[];
int v[];
class AC
{
private:
int ch[N][child_num];
int fail[N];
int Q[N];
int val[N];
int sz;
int id[];
int dp[][N][<<];
char s[N];
public:
void init()
{
fail[] = ;
id['A'] = ,id['G'] = ,id['T'] = ,id['C'] = ;
}
void reset()
{
memset(ch[],,sizeof(ch[]));
memset(val,,sizeof(val));
sz = ;
}
void insert(char *a,int key)
{
int p = ;
for(; *a ; a++)
{
int d= id[*a];
if(ch[p][d]==)
{
memset(ch[sz],,sizeof(ch[sz]));
s[sz] = *a;
ch[p][d] = sz++;
}
p = ch[p][d];
}
val[p] = (<<key);
}
void construct()
{
int i,head=,tail = ;
for(i = ; i < child_num ;i++)
{
if(ch[][i])
{
fail[ch[][i]] = ;
Q[tail++] = ch[][i];
}
}
while(head!=tail)
{
int u = Q[head++];
val[u]|=val[fail[u]];
for(i = ; i < child_num ; i++)
{
if(ch[u][i])
{
fail[ch[u][i]] = ch[fail[u]][i];
Q[tail++] = ch[u][i];
}
else ch[u][i] = ch[fail[u]][i];
}
}
}
void work(int n,int m)
{
int i,j,g;
memset(dp,,sizeof(dp));
dp[][][] = ;
for(i = ;i < n ;i++)
{
memset(dp[(i+)%],,sizeof(dp[(i+)%]));
for(j = ;j < sz ; j++)
{
for(int e = ; e < (<<m) ; e++)
{
if(!dp[i%][j][e]) continue;
for(g = ; g < child_num ; g++)
{
int o = val[ch[j][g]];
dp[(i+)%][ch[j][g]][e|o] = dp[i%][j][e];
}
}
}
}
int cnt = -INF;
for(i = ;i < sz ; i++)
{
for(j = ;j < (<<m) ; j++)
{
int ans = ;
if(!dp[n%][i][j]) continue;
for(g = ; g < m ;g++)
{
if(j&(<<g))
ans+=v[g];
}
cnt = max(ans,cnt);
}
}
if(cnt<)
puts("No Rabbit after 2012!");
else
printf("%d\n",cnt);
}
}ac;
int main()
{
int n,i,m;
ac.init();
while(scanf("%d%d",&m,&n)!=EOF)
{
ac.reset();
for(i = ;i <= m ;i++)
{
scanf("%s%d",vir,&v[i-]);
ac.insert(vir,i-);
}
ac.construct();
ac.work(n,m);
}
return ;
}
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