洛谷P4608 [FJOI2016]所有公共子序列问题 【序列自动机 + dp + 高精】

题目链接

洛谷P4608

题解

建个序列自动机后
第一问暴搜
第二问dp + 高精
设\(f[i][j]\)为两个序列自动机分别走到\(i\)和\(j\)节点的方案数，答案就是\(f[0][0]\)
由于空间卡的很紧，高精不仅要压位，还要动态开内存
由于有些状态是没用的，记忆化搜索以减少内存损失

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 3015,P = 1000000000,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
char X[maxn],Y[maxn];
int typ,n,m,last[60];
inline int id(char c){return c >= 'a' ? 26 + c - 'a' : c - 'A';}
struct LAM{
	int ch[maxn][52],cnt;
	void build(char* S,int len){
		for (int i = 0; i < 52; i++) last[i] = 0;
		cnt = len;
		for (int i = len; i; i--){
			for (int j = 0; j < 52; j++)
				ch[i][j] = last[j];
			last[id(S[i])] = i;
		}
		for (int i = 0; i < 52; i++) ch[0][i] = last[i];
	}
}A,B;
char s[maxn];
int len,ans;
void dfs(int u,int v){
	ans++;
	for (int i = 1; i <= len; i++) putchar(s[i]); puts("");
	for (int i = 0; i < 52; i++)
		if (A.ch[u][i] && B.ch[v][i]){
			s[++len] = i > 25 ? 'a' + i - 26 : 'A' + i;
			dfs(A.ch[u][i],B.ch[v][i]);
			len--;
		}
}
void work1(){
	dfs(0,0);
	printf("%d\n",ans);
};
struct NUM{
	int len;
	LL* s;
	void init(){
		s = new LL[20];
		for (int i = 0; i < 20; i++) s[i] = 0;
		len = 0;
	}
	void out(){
		if (!len){puts("0"); return;}
		printf("%lld",s[len - 1]);
		for (int i = len - 2; ~i; i--)
			printf("%09lld",s[i]);
	}
	void add(const NUM& a){
		LL carry = 0,tmp,L = max(len,a.len);
		for (int i = 0; i < L; i++){
			tmp = s[i] + a.s[i] + carry;
			s[i] = tmp % P;
			carry = tmp / P;
		}
		if (carry) s[L] += carry;
		len = 0;
		for (int i = 19; ~i; i--) if (s[i]){len = i + 1; break;}
	}
}f[maxn][maxn];
int vis[maxn][maxn];
void DFS(int u,int v){
	if (vis[u][v]) return;
	vis[u][v] = true;
	f[u][v].init();
	f[u][v].s[0] = f[u][v].len = 1;
	for (int i = 0; i < 52; i++)
		if (A.ch[u][i] && B.ch[v][i]){
			DFS(A.ch[u][i],B.ch[v][i]);
			f[u][v].add(f[A.ch[u][i]][B.ch[v][i]]);
		}
}
void work2(){
	DFS(0,0);
	f[0][0].out();
}
int main(){
	n = read(); m = read();
	scanf("%s%s%d",X + 1,Y + 1,&typ);
	A.build(X,n); B.build(Y,m);
	if (typ) work1();
	else work2();
	return 0;
}