제출 #764647

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
764647		GrindMachine	Tents (JOI18_tents)	C++17	100 / 100	97 ms	70948 KiB

tents

// Om Namah Shivaya

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

refs:
https://codeforces.com/blog/entry/58433?#comment-421099

no row/col has > 2 points

key idea:
order the points in some way
increasing coordinate of column (j value)

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

void solve(int test_case)
{
    ll n,m; cin >> n >> m;
    ll dp[n+5][m+5];
    memset(dp,0,sizeof dp);

    rep(i,n+1){
        dp[i][0] = 1;
    }
    rep(j,m+1){
        dp[0][j] = 1;
    }

    rep1(i,n){
        rep1(j,m){
            // place nothing on curr col
            dp[i][j] += dp[i][j-1];

            // place individual guy on curr col
            dp[i][j] += i*4*dp[i-1][j-1];

            // place pair on curr col
            dp[i][j] += (i*(i-1)/2) * dp[i-2][j-1];

            // place pair on same row, but one of the points must be in curr col
            // i ways to pick row
            // first point is in col j
            // j-1 ways to pick col of other point
            dp[i][j] += i*(j-1) * dp[i-1][j-2];
            
            dp[i][j] %= MOD;
        }
    }

    // sub 1 for empty case
    ll ans = (dp[n][m]-1+MOD) % MOD;
    cout << ans << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

• Subtask 1: 48 / 48
• Subtask 2: 52 / 52