HDU 3341 Lost's revenge（AC自动机+DP+变进制优化）

题目是给一个DNA重新排列使其包含最多的数论基因。

字符最长是40
只需要记录ACGT出现的次数。
如果使用5维数组，显然超内存了。
假设ACGT的总数分别为num[0],num[1],num[2],num[3]
那么对于ACGT的数量分别为ABCD的状态可以记录为：
A*(num[1]+1)*(num[2]+1)*(num[3]+1) + B*(num[2]+1)*(num[3]+1)+ C*(num[3]+1) +D
这样的状态最大就是11*11*11*11
复杂度也可以承受了。
字符串可能有重复的，用int来记录数量

忘了dp【j】【i】小于0直接continue的情况，wa了好多。。

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF  0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-9
#define maxn 60010
#define MOD 1000000007

int n,m;
int dp[510][20010];
int bit[13],num[13];
struct Trie
{
int next[510][5],fail[510],en[510];
int root,L;
void init()
{
L = 0;
root = newnode();
}
int newnode()
{
for(int i = 0; i < 4; i++)
next[L][i] = -1;
en[L++] = 0;
return L-1;
}
int getch(char ch)
{
if(ch == 'A')
return 0;
else if(ch == 'C')
return 1;
else if(ch == 'G')
return 2;
return 3;
}
void Insert(char buf[])
{
int now = root;
int len = strlen(buf);
for(int i = 0; i < len; i++)
{
if(next[now][getch(buf[i])] == -1)
next[now][getch(buf[i])] = newnode();
now = next[now][getch(buf[i])];
}
en[now]++;
}
void build()
{
queue<int> Q;
fail[root] = root;
for(int i = 0; i < 4; i++)
{
if(next[root][i] == -1)
next[root][i] = root;
else
{
fail[next[root][i]] = root;
Q.push(next[root][i]);
}
}
while(!Q.empty())
{
int now = Q.front();
Q.pop();
en[now] += en[fail[now]];
for(int i = 0; i < 4; i++)
{
if(next[now][i] == -1)
next[now][i] = next[fail[now]][i];
else
{
fail[next[now][i]] = next[fail[now]][i];
Q.push(next[now][i]);
}
}
}
}
int ok(int x,int &a,int &b,int &c,int &d)
{
a = x/bit[0];
if(a > num[0])
return 0;
x -= a*bit[0];
b = x/bit[1];
if(b > num[1])
return 0;
x -= b*bit[1];
c = x/bit[2];
if(c > num[2])
return 0;
x -= c*bit[2];
d = x/bit[3];
if(d > num[3] || x-d*bit[3] != 0)
return 0;

return 1;
}
int solve()
{
char s[1010];
scanf("%s",s);
int len = strlen(s);
memset(num,0,sizeof(num));
for(int i = 0; i < len; i++)
num[getch(s[i])]++;
bit[0] = (num[1]+1)*(num[2]+1)*(num[3]+1);
bit[1] = (num[2]+1)*(num[3]+1);
bit[2] = num[3]+1;
bit[3] = 1;
int ss = num[0]*bit[0] + num[1]*bit[1] + num[2]*bit[2] + num[3]*bit[3];
memset(dp,-1,sizeof(dp));
dp[root][0] = 0;
for(int i = 0; i <= ss; i++)
{
int a,b,c,d;
if(ok(i,a,b,c,d))
for(int j = 0; j < L; j++)
{
if(dp[j][i] < 0)
continue;
if(a < num[0])
dp[next[j][0]][i+bit[0]] = max(dp[next[j][0]][i+bit[0]],dp[j][i]+en[next[j][0]]);
if(b < num[1])
dp[next[j][1]][i+bit[1]] = max(dp[next[j][1]][i+bit[1]],dp[j][i]+en[next[j][1]]);
if(c < num[2])
dp[next[j][2]][i+bit[2]] = max(dp[next[j][2]][i+bit[2]],dp[j][i]+en[next[j][2]]);
if(d < num[3])
dp[next[j][3]][i+bit[3]] = max(dp[next[j][3]][i+bit[3]],dp[j][i]+en[next[j][3]]);
}
}
int ans = 0;
for(int i = 0; i < L; i++)
ans = max(ans,dp[i][ss]);
return ans;
}
} ac;
int main()
{
int t,C = 1;
//scanf("%d",&t);
while(scanf("%d",&n) && n)
{
ac.init();
char s[1010];
for(int i = 0; i < n; i++)
{
scanf("%s",s);
ac.Insert(s);
}
ac.build();
printf("Case %d: %d\n",C++,ac.solve());
}
return 0;
}

大牛代码：

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF  0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-9
#define maxn 60010
#define MOD 1000000007

int n,m;
int dp[510][20010];
int bit[13],num[13];
struct Trie
{
int next[510][5],fail[510],en[510];
int root,L;
void init()
{
L = 0;
root = newnode();
}
int newnode()
{
for(int i = 0; i < 4; i++)
next[L][i] = -1;
en[L++] = 0;
return L-1;
}
int getch(char ch)
{
if(ch == 'A')
return 0;
else if(ch == 'C')
return 1;
else if(ch == 'G')
return 2;
return 3;
}
void Insert(char buf[])
{
int now = root;
int len = strlen(buf);
for(int i = 0; i < len; i++)
{
if(next[now][getch(buf[i])] == -1)
next[now][getch(buf[i])] = newnode();
now = next[now][getch(buf[i])];
}
en[now]++;
}
void build()
{
queue<int> Q;
fail[root] = root;
for(int i = 0; i < 4; i++)
{
if(next[root][i] == -1)
next[root][i] = root;
else
{
fail[next[root][i]] = root;
Q.push(next[root][i]);
}
}
while(!Q.empty())
{
int now = Q.front();
Q.pop();
en[now] += en[fail[now]];
for(int i = 0; i < 4; i++)
{
if(next[now][i] == -1)
next[now][i] = next[fail[now]][i];
else
{
fail[next[now][i]] = next[fail[now]][i];
Q.push(next[now][i]);
}
}
}
}
int solve()
{
char s[1010];
scanf("%s",s);
int len = strlen(s);
memset(num,0,sizeof(num));
for(int i = 0; i < len; i++)
num[getch(s[i])]++;
bit[0] = (num[1]+1)*(num[2]+1)*(num[3]+1);
bit[1] = (num[2]+1)*(num[3]+1);
bit[2] = num[3]+1;
bit[3] = 1;
memset(dp,-1,sizeof(dp));
dp[root][0] = 0;
for(int A = 0; A <= num[0]; A++)
for(int B = 0; B <= num[1]; B++)
for(int C = 0; C <= num[2]; C++)
for(int D = 0; D <= num[3]; D++)
{
int s = A*bit[0] + B*bit[1] + C*bit[2] + D*bit[3];
for(int i = 0; i < L; i++)
if(dp[i][s] >= 0)
{
for(int k = 0; k < 4; k++)
{
if(k == 0 && A == num[0])
continue;
if(k == 1 && B == num[1])
continue;
if(k == 2 && C == num[2])
continue;
if(k == 3 && D == num[3])
continue;
dp[next[i][k]][s+bit[k]] = max(dp[next[i][k]][s+bit[k]],dp[i][s]+en[next[i][k]]);
}
}
}
int ans = 0;
int ss = num[0]*bit[0] + num[1]*bit[1] + num[2]*bit[2] + num[3]*bit[3];
for(int i = 0; i < L; i++)
ans = max(ans,dp[i][ss]);
return ans;
}
} ac;
int main()
{
int t,C = 1;
//scanf("%d",&t);
while(scanf("%d",&n) && n)
{
ac.init();
char s[1010];
for(int i = 0; i < n; i++)
{
scanf("%s",s);
ac.Insert(s);
}
ac.build();
printf("Case %d: %d\n",C++,ac.solve());
}
return 0;
}