Nowcoder Typing practice ( Trie 图 )

题目链接

题意 : 给出 n 个串、然后给出一个问询串、问你对于问询串的每一个前缀、需要至少补充多少单词才能使得其后缀包含 n 个串中的其中一个、注意 '-' 字符代表退格

 

分析 :

多串的匹配问询自然想到 AC 自动机 或者 构建 Trie 图

首先将 N 个串丢到 Trie 图里面

对于每一个节点记录其要变成一个完整的串最少需要补充的单词数

然后在问询的时候、由于有退格操作

于是需要将跑过的节点路径记录下来以便恢复

这个我们可以使用栈来做到

然后对于问询串的每一个前缀问询

可以采用 DP 的方式来一直记录跳 Fail 时候每个节点的最优值是什么

 

#include<bits/stdc++.h>
#define LL long long
#define ULL unsigned long long

#define scl(i) scanf("%lld", &i)
#define scll(i, j) scanf("%lld %lld", &i, &j)
#define sclll(i, j, k) scanf("%lld %lld %lld", &i, &j, &k)
#define scllll(i, j, k, l) scanf("%lld %lld %lld %lld", &i, &j, &k, &l)

#define scs(i) scanf("%s", i)
#define sci(i) scanf("%d", &i)
#define scd(i) scanf("%lf", &i)
#define scIl(i) scanf("%I64d", &i)
#define scii(i, j) scanf("%d %d", &i, &j)
#define scdd(i, j) scanf("%lf %lf", &i, &j)
#define scIll(i, j) scanf("%I64d %I64d", &i, &j)
#define sciii(i, j, k) scanf("%d %d %d", &i, &j, &k)
#define scddd(i, j, k) scanf("%lf %lf %lf", &i, &j, &k)
#define scIlll(i, j, k) scanf("%I64d %I64d %I64d", &i, &j, &k)
#define sciiii(i, j, k, l) scanf("%d %d %d %d", &i, &j, &k, &l)
#define scdddd(i, j, k, l) scanf("%lf %lf %lf %lf", &i, &j, &k, &l)
#define scIllll(i, j, k, l) scanf("%I64d %I64d %I64d %I64d", &i, &j, &k, &l)

#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define lowbit(i) (i & (-i))
#define mem(i, j) memset(i, j, sizeof(i))

#define fir first
#define sec second
#define VI vector<int>
#define ins(i) insert(i)
#define pb(i) push_back(i)
#define pii pair<int, int>
#define VL vector<long long>
#define mk(i, j) make_pair(i, j)
#define all(i) i.begin(), i.end()
#define pll pair<long long, long long>

#define _TIME 0
#define _INPUT 0
#define _OUTPUT 0
clock_t START, END;
void __stTIME();
void __enTIME();
void __IOPUT();
using namespace std;

const int max_node = 1e6 + 10;
const int max_len = 1e5 + 10;
const int Letter  = 26;

struct Aho{
    struct StateTable{
        int nxt[Letter];
        int fail, cnt, dis;
        bool vis;
        void init(){
            memset(nxt, 0, sizeof(nxt));
            fail = 0;
            cnt = 0;
            dis = 0x3f3f3f3f;
            vis = false;
        }
    }Node[max_node];

    int sz;
    queue<int> que;

    inline void init(){ while(!que.empty())que.pop(); Node[0].init(); sz = 1; }

    inline void insert(char *s, int len){
        int now = 0;
        Node[now].dis = min(Node[now].dis, len);
        for(int i=0; i<len; i++){
            int idx = s[i] - 'a';
            if(!Node[now].nxt[idx]){
                Node[sz].init();
                Node[now].nxt[idx] = sz++;
            }
            now = Node[now].nxt[idx];
            Node[now].dis = min(Node[now].dis, len-i-1);
        }
        Node[now].cnt++;
    }

    inline void build(){
        Node[0].fail = -1;
        que.push(0);
        while(!que.empty()){
            int top = que.front();  que.pop();
            for(int i=0; i<Letter; i++){
                if(Node[top].nxt[i]){
                    if(top == 0) Node[ Node[top].nxt[i] ].fail = 0;
                    else{
                        int v = Node[top].fail;
                        while(v != -1){
                            if(Node[v].nxt[i]){
                                Node[ Node[top].nxt[i] ].fail = Node[v].nxt[i];
                                break;
                            }v = Node[v].fail;
                        }if(v == -1) Node[ Node[top].nxt[i] ].fail = 0;
                    }que.push(Node[top].nxt[i]);
                }else Node[top].nxt[i] = top!=0?Node[ Node[top].fail ].nxt[i]:0;
            }
        }
    }

//    int Match(char *s){
//        int now = 0, res = 0;
//        for(int i=0; s[i]!='\0'; i++){
//            int idx = s[i] - 'a';
//            now = Node[now].nxt[idx];
//            int tmp = now;
//            while(tmp != 0){
//                res += Node[tmp].cnt;
//                Node[tmp].cnt = 0;
//                tmp = Node[tmp].fail;
//            }
//        }
//        return res;
//    }

    int dp[max_node];
    int GetDP(int cur){
        if(cur == 0) return Node[cur].dis;
        if(dp[cur] != -1) return dp[cur];
        dp[cur] = min(Node[cur].dis, GetDP(Node[cur].fail));
        return dp[cur];
    }
    void query(char *s, int len){
        int now = 0;
        memset(dp, -1, sizeof(dp));
        stack<int> pos;
        printf("%d\n", Node[now].dis);
        for(int i=0; i<len; i++){
            if(s[i] == '-'){
                if(!pos.empty()) pos.pop();
                if(pos.empty()) now = 0;
                else now = pos.top();
            }else{
                int idx = s[i] - 'a';
                now = Node[now].nxt[idx];
                pos.push(now);
            }

            int ans = Node[now].dis;
            ans = min(ans, GetDP(now));

            printf("%d\n", ans);
        }
    }
}ac;

char s[max_len];
int main(void){__stTIME();__IOPUT();

    int n;

    sci(n);

    ac.init();

    for(int i=0; i<n; i++){
        scs(s);
        ac.insert(s, strlen(s));
    }

    ac.build();

    scs(s);

    ac.query(s, strlen(s));

__enTIME();return 0;}


void __stTIME()
{
    #if _TIME
        START = clock();
    #endif
}

void __enTIME()
{
    #if _TIME
        END = clock();
        cerr<<"execute time = "<<(double)(END-START)/CLOCKS_PER_SEC<<endl;
    #endif
}

void __IOPUT()
{
    #if _INPUT
        freopen("in.txt", "r", stdin);
    #endif
    #if _OUTPUT
        freopen("out.txt", "w", stdout);
    #endif
}