제출 #1017366

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1017366		caterpillow	Matching (CEOI11_mat)	C++17	0 / 100	2094 ms	50152 KiB

mat

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using pl = pair<ll, ll>;
using pi = pair<int, int>;
#define vt vector
#define f first
#define s second
#define pb push_back
#define all(x) x.begin(), x.end() 
#define size(x) ((int) (x).size())
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define F0R(i, b) FOR (i, 0, b)
#define endl '\n'
const ll INF = 1e18;
const int inf = 1e9;

template<template<typename> class Container, typename T>
ostream& operator<<(ostream& os, Container<T> o) {
    os << "{"; 
    int g = size(o); 
    for (auto i : o) os << i << ((--g) == 0 ? "" : ", "); 
    os << "}";
    return os;
}

void _print() {
    cerr << "\n";
}

template<typename T, typename ...V>
void _print(T t, V... v) {
    cerr << t; if (sizeof...(v)) cerr << ", "; _print(v...);
}

#ifdef LOCAL
#define dbg(x...) cerr << #x << " = "; _print(x);
#else
#define dbg(x...)
#define cerr if (0) std::cerr
#endif

/*

try sus kmp?

precompute the indices i and j for each k in the pattern 
to quickly check is some building can be glued to the end of the current match
a match is valid if and only if this is satisfied for all buildings

compute a fallback array, such that if a mismatch happens on index i you can delete
the first x amount to make the prefix match again

compute an auxilliary array dp, where dp[i] = how far away the next mismatch is 
    if you were to start matching at index i

if you know that from index i, the first mismatched character is j, and that 
if you were to match with yourself from index 1 the next mismatched character is at 1 + k, 
then
    case 1: at index i + 1, you know for certain that index i + 1 + k is a mismatch, and i + 1 + k < j
        this means that you know the next mismatched position
        you can continue doing this until case 2
    case 2: at index i + 1, you know for certain that index i + 1 + k is a mismatch, and i + 1 + k >= j
        who knows, it might match now!
        start checking again from index i + 1

*/

int m, n;
vt<int> pat;
vt<pi> comp;
vt<int> h;

// check match at index j of the pattern starting at index i in str
bool match(int i, int j, vt<int>& str) { 
    // cerr << "matching " << j << " starting from " << i << endl;
    pi prv = comp[j - i];
    bool good = true;
    if (prv.f != -1) good &= str[prv.f + i] < str[j];
    if (prv.s != -1) good &= str[j] < str[prv.s + i];
    return good;
}

vt<int> dp;
void compute_dp() {
    dp.resize(m, inf);
    dp[0] = 0;
    int j = 1; // the first mismatched character starting from index i
    FOR (i, 1, m) {
        while (j < m && match(i, j, pat)) j++;
        cerr << "matched from " << i << " up to " << j << endl;
        dp[i] = j - i;
        cerr << "dp " << i << " = " << j - i << endl;
        int k = 1;
        cerr << i + k << " " << (i + k < m ? i + k + dp[k] : -1) << endl;
        while (i + k < m && i + k + dp[k] < j) { // case 1: we know the mismatch happens in our good prefix
            cerr << "we know the next mismatch\n";
            dp[i + k] = dp[k];
            cerr << "dp " << i + k << " = " << dp[k] << endl;
            k++;
        }
        i += k - 1;
        // case 2: idk whats happenign anymore
    }
    dbg(dp);
}   

vt<int> fb;
void compute_fallback() {
    fb.resize(m + 1);
    fb[0] = -1;
    vt<int> mod_dp = dp; // convert from next mismatch to last matching index relative to start
    mod_dp[0] = 1;
    dbg(mod_dp);
    vt<int> last(m + 1, m);
    ROF (i, 0, m) {
        fb[i + 1] = last[mod_dp[i]] - i;
        last[mod_dp[i]] = i;
        dbg(last);
    }
    dbg(fb);
}

main() {
    cin.tie(0)->sync_with_stdio(0);
    
    cin >> m >> n;
    pat.resize(m);
    F0R (i, m) {
        int v; cin >> v;
        pat[v - 1] = i;
    }
    comp.resize(m, {-1, -1});
    map<int, int> prv;
    F0R (i, m) {
        auto it = prv.lower_bound(pat[i]);
        if (it != prv.end()) comp[i].s = it->s;
        if (it != prv.begin()) comp[i].f = prev(it)->s;
        prv[pat[i]] = i;
    }
    h.resize(n);
    F0R (i, n) cin >> h[i];

    dbg(pat);
    F0R (i, m) {
        cerr << comp[i].f << " " << comp[i].s << endl; 
    }

    compute_dp();
    compute_fallback();

    vt<int> out;
    int i = 0, j = 0;
    while (i < n) {
        while (j < n && j - i < m && match(i, j, h)) j++;
        cerr << "starting from " << i << ", mismatched at " << j << endl;
        // j is now at first mismatch
        if (j - i == m) out.pb(i);
        i += fb[j - i];
        cerr << "i = " << i << endl;
    }
    cout << size(out) << endl;
    for (int e : out) cout << e + 1 << " ";
    cout << endl;
}

컴파일 시 표준 에러 (stderr) 메시지

mat.cpp:127:1: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
  127 | main() {
      | ^~~~

• Subtask 1: 0 / 9
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