2019 ICPC Asia Nanjing Regional F. Paper Grading

Posted 2020-11-24 zeronera

题目链接：https://nanti.jisuanke.com/t/42400

题意：给一个模式串集合，序号1~n，有T个操作，或者是交换序号，或者是查询模式串集合中序号在L到R之 间的字符串有多少个和目标串公共前缀长度大于等于K。
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对模式串集合建字典树，则与目标串(LCP)大于等K的字符串，都在以目标串第(K)个字符所在的结点为根的子树中，每个模式串插入字典树的过程中记录序号，修改时交换序号，询问就相于在子树中询问有多少结点的序号在(L)到(R)之间。
涉及到子树，容易想到建立(DFS)序，再用数据结构维护
但是不建议在字典树的(DFS)序上用数据结构维护，因为可能出现两个一样的模式串，但是序号不一样，可能会覆盖之前的序号（个人观点，我写的时候是遇到了这种问题，可能写法问题）
所以可以用一个数组(h[i])，表示序号为(i)的字符串在字典树中的编号为(h[i])，对(h)使用数据结构维护，需要支持单点修改（交换视为两次单点修改）和查询区间([L,R])中(h[i])的值在范围([in[x],out[x]])中的数量（(x)为目标串第(K)字符在字典树中的结点编号）

#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
int ptov[maxn], tv[maxn];
/************************************************************/
struct Trie
{
    int trie[maxn][26], tot = 0;
    int ins(char s[])
    {
        int len = strlen(s);
        int root = 0;
        for (int i = 0; i < len; i++)
        {
            int id = s[i] - 'a';
            if (trie[root][id] == 0)
                trie[root][id] = ++tot;
            root = trie[root][id];
        }
        return root;
    }
    int fin(int k, char s[])
    {
        int root = 0;
        for (int i = 0; i < k; i++)
        {
            int id = s[i] - 'a';
            if (trie[root][id] == 0)
                return -1;
            root = trie[root][id];
        }
        return root;
    }
} t;
/************************************************************/
int in[maxn], out[maxn];
int tim = 0;
void dfs(int x)
{
    in[x] = ++tim;
    for (int i = 0; i < 26; i++)
    {
        if (t.trie[x][i] == 0)
            continue;
        dfs(t.trie[x][i]);
    }
    out[x] = tim;
}
/************************************************************/
int T[maxn], S[maxn], L[maxn * 150], R[maxn * 150], sum[maxn * 150];
int h[maxn];
int ul[maxn], ur[maxn];
int tot, num, n, m;

struct node
{
    int l, r, k;
    bool flag;
} Q[maxn << 1];

void build(int &rt, int l, int r)
{
    rt = ++tot;
    sum[rt] = 0;
    if (l == r)
        return;
    int mid = (l + r) >> 1;
    build(L[rt], l, mid);
    build(R[rt], mid + 1, r);
}

void update(int &rt, int pre, int l, int r, int x, int val)
{
    rt = ++tot;
    L[rt] = L[pre];
    R[rt] = R[pre];
    sum[rt] = sum[pre] + val;
    if (l == r)
        return;
    int mid = (l + r) >> 1;
    if (x <= mid)
        update(L[rt], L[pre], l, mid, x, val);
    else
        update(R[rt], R[pre], mid + 1, r, x, val);
}

int lowbit(int x) { return x & (-x); }

void add(int x, int val)
{
    int res = lower_bound(h + 1, h + 1 + num, ptov[x]) - h;
    while (x <= n)
    {
        update(S[x], S[x], 1, num, res, val);
        x += lowbit(x);
    }
}

int Sum(int x, int flag)
{
    int res = 0;
    while (x > 0)
    {
        if (flag == 1)
            res += sum[L[ur[x]]];
        else if (flag == 2)
            res += sum[L[ul[x]]];
        else if (flag == 3)
            res += sum[ur[x]];
        else
            res += sum[ul[x]];
        x -= lowbit(x);
    }
    return res;
}
int query(int s, int e, int ts, int te, int l, int r, int k)
{
    if (l == r)
        return Sum(e, 3) - Sum(s, 4) + sum[te] - sum[ts];
    int mid = (l + r) >> 1;
    int res = Sum(e, 1) - Sum(s, 2) + sum[L[te]] - sum[L[ts]];
    if (k <= mid)
    {
        for (int i = e; i; i -= lowbit(i))
            ur[i] = L[ur[i]];
        for (int i = s; i; i -= lowbit(i))
            ul[i] = L[ul[i]];
        return query(s, e, L[ts], L[te], l, mid, k);
    }
    else
    {
        for (int i = e; i; i -= lowbit(i))
            ur[i] = R[ur[i]];
        for (int i = s; i; i -= lowbit(i))
            ul[i] = R[ul[i]];
        return res + query(s, e, R[ts], R[te], mid + 1, r, k);
    }
}
/************************************************************/
char s[maxn];
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
    {
        scanf("%s", s);
        ptov[i] = t.ins(s);
    }
    dfs(0);
    for (int i = 1; i <= n; i++)
    {
        ptov[i] = in[ptov[i]];
        h[i] = ptov[i];
        tv[i] = ptov[i];
    }
    int q = 0;
    for (int i = 1; i <= m; i++)
    {
        int opt;
        scanf("%d", &opt);
        if (opt == 2) //询问
        {
            scanf("%s%d%d%d", s, &Q[q].k, &Q[q].l, &Q[q].r);
            Q[q].k = t.fin(Q[q].k, s);
            Q[q].flag = true;
            q++;
        }
        else //修改
        {
            int tl, tr;
            scanf("%d%d", &tl, &tr);
            Q[q].l = tl;
            Q[q].r = tv[tr];
            Q[q].flag = false;
            q++;

            Q[q].l = tr;
            Q[q].r = tv[tl];
            Q[q].flag = false;
            q++;

            swap(tv[tl], tv[tr]);
        }
    }
    sort(h + 1, h + 1 + n);
    num = unique(h + 1, h + 1 + n) - h - 1;
    tot = 0;
    build(T[0], 1, num);
    for (int i = 1; i <= n; i++)
        update(T[i], T[i - 1], 1, num, lower_bound(h + 1, h + 1 + num, ptov[i]) - h, 1);
    for (int i = 1; i <= n; i++)
        S[i] = T[0];
    for (int i = 0; i < q; i++)
    {
        if (Q[i].flag)
        {
            if (Q[i].k == -1)
                puts("0");
            else if (Q[i].k == 0)
                printf("%d
", Q[i].r - Q[i].l + 1);
            else
            {
                int tin = lower_bound(h + 1, h + 1 + num, in[Q[i].k]) - h;
                auto tout = lower_bound(h + 1, h + 1 + num, out[Q[i].k]);
                for (int j = Q[i].r; j; j -= lowbit(j))
                    ur[j] = S[j];
                for (int j = Q[i].l - 1; j; j -= lowbit(j))
                    ul[j] = S[j];
                int ans1 = query(Q[i].l - 1, Q[i].r, T[Q[i].l - 1], T[Q[i].r], 1, num, tout - h);
                if (tin == 1)
                {
                    if (*tout == out[Q[i].k])
                        printf("%d
", ans1);
                    else
                        printf("%d
", ans1 - 1);
                    continue;
                }
                for (int j = Q[i].r; j; j -= lowbit(j))
                    ur[j] = S[j];
                for (int j = Q[i].l - 1; j; j -= lowbit(j))
                    ul[j] = S[j];
                int ans2 = query(Q[i].l - 1, Q[i].r, T[Q[i].l - 1], T[Q[i].r], 1, num, tin - 1);
                if (*tout == out[Q[i].k])
                    printf("%d
", ans1 - ans2);
                else
                    printf("%d
", ans1 - 1 - ans2);
            }
        }
        else
        {
            add(Q[i].l, -1);
            ptov[Q[i].l] = Q[i].r;
            add(Q[i].l, 1);
        }
    }
    return 0;
}