HDU 3341 Lost's revenge (AC自动机+DP)

题意：给定 n 个子串，然后给一个母串，让你对母串重排，然后问你最多含有多少个字串。

析：这个题很容易想到五维DP，这样的话，不但会MLE，而且连数组都不一定开的出来，里面有大量的无用的状态，所以我们把那四个字符出现的次数进行重新编制，假设A出现 a 次，C出现 c 次，G出现 g 次，T出现 t 次，进行压缩的时候，把第A的系数乘以((b+1) + (c+1) + (d+1))，C乘以((c+1) + (d+1))，G乘以(d+1)，T乘以1。这样就很简单的dp[i][j] 表示在 i 这个结点 j 状态时，最多有多少个。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
//#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 11*11*11*11 + 10;
const int maxm = 1e5 + 10;
const int mod = 50007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
const int maxnode = 50 * 10 + 10;
const int sigma = 4;

int dp[maxnode][maxn];

struct Aho{
  int ch[maxnode][sigma];
  int f[maxnode];
  int val[maxnode];
  int sz;

  void clear(){ sz = 1;  ms(ch[0], 0); }
  inline int idx(char ch){
    if(ch == 'A')  return 0;
    if(ch == 'C')  return 1;
    if(ch == 'G')  return 2;
    return 3;
  }

  void insert(const char *s){
    int u = 0;
    for(int i = 0; s[i]; ++i){
      int c = idx(s[i]);
      if(!ch[u][c]){
        ms(ch[sz], 0);
        val[sz] = 0;
        ch[u][c] = sz++;
      }
      u = ch[u][c];
    }
    ++val[u];
  }

  void getFail(){
    queue<int> q;  f[0] = 0;
    for(int i = 0; i < sigma; ++i){
      int u = ch[0][i];
      if(u){ f[u] = 0;  q.push(u); }
    }
    while(!q.empty()){
      int r = q.front();  q.pop();
      for(int c = 0; c < sigma; ++c){
        int u = ch[r][c];
        if(!u){ ch[r][c] = ch[f[r]][c];  continue; }
        q.push(u);
        int v = f[r];
        while(v && !ch[v][c])  v = f[v];
        f[u] = ch[v][c];
        val[u] += val[f[u]];
      }
    }
  }

  int solve(int s1, int s2, int s3, int s4){
    ms(dp, -1);  dp[0][0] = 0;
    int cnt1 = s2 * s3 * s4;
    int cnt2 = s3 * s4;
    int cnt3 = s4;
    int ans = 0;
    FOR(j, 0, s1)  FOR(k, 0, s2)
    FOR(l, 0, s3)  FOR(y, 0, s4) FOR(i, 0, sz){
      int pre = j * cnt1 + k * cnt2 + l * cnt3 + y;
      if(dp[i][pre] < 0)  continue;
      if(j + 1 < s1){
        int nxt = ch[i][0];
        int st = pre + cnt1;
        dp[nxt][st] = max(dp[nxt][st], dp[i][pre] + val[nxt]);
        ans = max(ans, dp[nxt][st]);
      }
      if(k + 1 < s2){
        int nxt = ch[i][1];
        int st = pre + cnt2;
        dp[nxt][st] = max(dp[nxt][st], dp[i][pre] + val[nxt]);
        ans = max(ans, dp[nxt][st]);
      }
      if(l + 1 < s3){
        int nxt = ch[i][2];
        int st = pre + cnt3;
        dp[nxt][st] = max(dp[nxt][st], dp[i][pre] + val[nxt]);
        ans = max(ans, dp[nxt][st]);
      }
      if(y + 1 < s4){
        int nxt = ch[i][3];
        int st = pre + 1;
        dp[nxt][st] = max(dp[nxt][st], dp[i][pre] + val[nxt]);
        ans = max(ans, dp[nxt][st]);
      }
    }
    return ans;
  }
};

Aho aho;
char s[50];

int main(){
  int kase = 0;
  while(scanf("%d", &n) == 1 && n){
    aho.cl;
    for(int i = 0; i < n; ++i){
      scanf("%s", s);
      aho.insert(s);
    }
    aho.getFail();
    scanf("%s", s);
    int cnt[4] = {1, 1, 1, 1};
    for(int i = 0; s[i]; ++i)
      ++cnt[aho.idx(s[i])];
    printf("Case %d: %d
", ++kase, aho.solve(cnt[0], cnt[1], cnt[2], cnt[3]));
  }
  return 0;
}