bzoj 5337 [TJOI2018] str

Link

Solution

水題

直接 \(f[i][j]\) 表示以第 \(i\) 位為結束位置，當前已經匹配了前 \(j\) 個氨基酸的方案數

使用雜湊轉移

轉移複雜度 \(O(10)\)，總複雜度 \(1e7\)

Code

#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<iostream>
#include<queue>
#include<string>
#include<ctime>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<long long,long long> pll;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define rep(i,j,k)  for(register int i=(int)(j);i<=(int)(k);i++)
#define rrep(i,j,k) for(register int i=(int)(j);i>=(int)(k);i--)
#define Debug(...) fprintf(stderr, __VA_ARGS__)

ll read(){
    ll x=0,f=1;char c=getchar();
    while(c<'0' || c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0' && c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

const int md = 1e9 + 7;
const ll mod = 2002070919ll;
const int maxn = 10010;
int n, len;
char s[maxn], t[maxn];
int l[110][12], a[110], f[maxn][110];
ll hsh[110][12], pw[maxn], h[maxn];

inline void add(int &x, int y) {
    x += y;
    if (x >= md) x -= md;
}
inline ll calc(int l, int r) {
    return (h[r] - h[l - 1] * pw[r - l + 1] % mod + mod) % mod;
}
void work(){
    n = read();
    scanf("%s", s + 1);
    len = strlen(s + 1);
    pw[0] = 1; rep(i, 1, len) pw[i] = pw[i - 1] * 100 % mod;
    rep(i, 1, len) h[i] = (h[i - 1] * 100 + s[i] - 'A') % mod;
    rep(i, 1, n) {
        a[i] = read();
        rep(j, 1, a[i]) {
            scanf("%s", t + 1);
            l[i][j] = strlen(t + 1);
            rep(k, 1, l[i][j]) hsh[i][j] = (hsh[i][j] * 100 + t[k] - 'A') % mod;
        }
    }
    rep(i, 0, len) f[i][0] = 1;
    rep(i, 1, len) rep(j, 1, n) {
        rep(k, 1, a[j]) {
            int nw = l[j][k];
            if (i < nw || !f[i - nw][j - 1]) continue;
            ll val = calc(i - nw + 1, i);
            if (val == hsh[j][k]) add(f[i][j], f[i - nw][j - 1]);
        }
    }
    int ans = 0;
    rep(i, 1, len) add(ans, f[i][n]);
    printf("%d\n", ans);
}

int main(){
    #ifdef LZT
        freopen("in","r",stdin);
    #endif
    
    work();
    
    #ifdef LZT
        Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);
    #endif
}
 

Review

注意雜湊的模數不能大過 \(INT\_MAX\)，不然兩個雜湊值相乘會爆 \(\text{long long}\)