oj.uz

답안 #374587

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
374587 2021-03-07T13:42:28 Z Karliver 건포도 (IOI09_raisins) C++14
100 / 100
 1297 ms 26992 KB 

#include <bits/stdc++.h>
#include <fstream>
#define FIXED_FLOAT(x)  std::fixed <<std::setprecision(3)<<(x)
#define all(v) (v).begin(), (v).end()
using namespace  std;
#define forn(i,n) for (int i = 0; i < (n); ++i)
using ll = long long;
int mod = (ll)1e9 + 7;
#define PI acos(-1)
typedef pair<int, int> pairs;
typedef complex<ll> G;
const int INF = 1e9 + 1;
const int N = 51;
const double eps = 1e-3;

template <typename XPAX>
void ckma(XPAX &x, XPAX y) {
    x = (x < y ? y : x);
}
template <typename XPAX>
void ckmi(XPAX &x, XPAX y) {
    x = (x > y ? y : x);
}


ll power(ll a, ll b){
    if(!b)
        return 1;
    ll c=power(a,b/2);
    c=(1LL*c*c);
    if(b%2)
        c=(1LL*c*a);
    return c;
}


int mul(int a, int b) {
    return a * 1ll * b % mod;
}
int add(int a, int b) {
    int s = (a+b);
    if (s>=mod) s-=mod;
    return s;
}



struct RMQ {
    vector<vector<int>>st;
    RMQ(vector<int> &a) {
        int n = a.size();
        int k = ceil(log2(n));
        st.clear();
        st = vector<vector<int>>(n, vector<int>(k + 1, INF));
        for(int i = 0;i < n;++i) {
            st[i][0] = a[i];
        }
        for(int j = 1;j <= k;++j) {
            for(int i = 0;i + (1 << j) <= n;++i) {
                st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
            }
        }

    }
    int rng_q(int l, int r) {
        //if(l > r)return -1;
        int j = log2(r - l + 1);
        int ans = min(st[l][j], st[r - (1 << j) + 1][j]);
        return ans;
    }

};

template<long long mod = 1000000007>
struct modint{
	long long a;

	modint() : a(0){}
	modint(long long t){
		a = t % mod;
		if(a < 0) a += mod;
	}

	operator long long() const{ return a; }

	bool operator==(const modint &x) const{ return a == x.a; }
	bool operator!=(const modint &x) const{ return a != x.a; }

	modint operator-() const{ return modint(a ? (mod - a) : 0); }
	modint operator~() const{ return pow(mod - 2); }

	modint operator+(const modint &x) const{ return modint(*this) += x; }
	modint operator-(const modint &x) const{ return modint(*this) -= x; }
	modint operator*(const modint &x) const{ return modint(*this) *= x; }
	modint operator/(const modint &x) const{ return modint(*this) /= x; }

	modint &operator+=(const modint &x){
		a += x.a;
		if(a >= mod) a -= mod;
		return *this;
	}
	modint &operator-=(const modint &x){
		a -= x.a;
		if(a < 0) a += mod;
		return *this;
	}
	modint &operator*=(const modint &x){
		a = a * x.a % mod;
		return *this;
	}
	modint &operator/=(const modint &x){
		a = a * (~x).a % mod;
		return *this;
	}

	friend ostream &operator<<(ostream &os,const modint &x){
		return os << x.a;
	}
	friend istream &operator>>(istream &is,modint &x){
		long long t;
		is >> t;
		x = modint(t);
		return is;
	}

	modint pow(long long x) const{
		modint ret = 1,tmp = a;
		for(;x;tmp *= tmp,x >>= 1){
			if(x & 1) ret *= tmp;
		}
		return ret;
	}
};
const ll MOD = 1e9 + 7;
using Mint = modint<MOD>;


template<class T>
struct Combination{
	vector<T> fact,factinv;
	Combination(int n) : fact(n + 1),factinv(n + 1){
		fact[0] = 1;
		for(int i = 1;i <= n;i++) fact[i] = fact[i - 1] * T(i);
		factinv[n] = ~fact[n];
		for(int i = n;i >= 1;i--) factinv[i - 1] = factinv[i] * T(i);
	}
	T nCr(int n,int r){
		if(n < 0 || r < 0 || n < r) return 0;
		return fact[n]  * factinv[r] *  factinv[n - r];
	}
	T factorial(int x) {
        if(x < 0)return 0;
        return fact[x];
	}
};


void done() {

    // -1
    // 1 -2
    // -2

}

void solve()
{


    // 4 5 2 3 1
    // 1 -3 1 -2
    // 4 6 3 1 5 2
    // 2 -3 -2 4 -3

    int n, m;
    cin >> n >> m;

    int DP[N][N][N][N];
    forn(i, N) forn(j, N) forn(l, N) forn(K, N) DP[i][j][l][K] = -1;
    int pr[N][N];
    forn(i, N)forn(j, N)pr[i][j] = 0;

    vector<vector<int>> a(n + 1, vector<int>(m + 1));
    for(int i = 1;i <= n;++i) {
        for(int j = 1;j <= m;++j) {
            cin >> a[i][j];
            pr[i][j] = pr[i - 1][j] + pr[i][j - 1] - pr[i - 1][j - 1] + a[i][j];
        }
    }

    function<int(int, int, int, int)> dfs = [&](int x1, int y1, int x2, int y2) {
        if (DP[x1][y1][x2][y2] > -1)
            return DP[x1][y1][x2][y2];
        if (x1 == x2 && y1 == y2)
        {
            DP[x1][y1][x2][y2] = 0;
            return 0;
        }
        int res = INT_MAX;
        int n1 = 0;
        int n2 = 0;
        for (int i = x1; i < x2; i++)
        {
            n1 = dfs(x1,y1,i,y2);
            n2 = dfs(i + 1,y1,x2,y2);
            if (n1 + n2 < res)
                res = n1 + n2;
        }
        for (int i = y1; i < y2; i++)
        {
            n1 = dfs(x1,y1,x2,i);
            n2 = dfs(x1,i + 1,x2,y2);
            if (n1 + n2 < res)
                res = n1 + n2;
        }
        res += pr[x2][y2] - pr[x1 - 1][y2] - pr[x2][y1 - 1] + pr[x1 - 1][y1 - 1];
        DP[x1][y1][x2][y2] = res;
        return res;

    };

    cout << dfs(1, 1, n, m) << '\n';


}


void emer() {





}

void another() {

    // 10011
    //
    int n, q;
    cin>> n >> q;
    vector<int> c(n);
    for(int i = 0;i < n;++i) {
        cin >> c[i];
    }

    vector<vector<pair<int, pairs>>> qur(n);
    for(int i = 0;i < q;++i) {
        int l, r;
        cin>> l >> r;
        --l;
        --r;
        qur[r].push_back({l, {i, 1}});
        if(l) {
            qur[l - 1].push_back({l, {i, -1}});
        }
    }
    vector<int> f(n);

    auto inc = [&](int x) {
        for(; x < n;x = (x | (x + 1))) f[x]++;
    };
    auto get = [&](int  x) {
        int ans =0;
        for(; x >= 0;x = (x & (x + 1)) - 1)
            ans += f[x];
        return ans;
    };

    vector<int> ret(q);
    map<int, int> last;
    vector<int> pev(n, -1);

    for(int i = 0;i < n;++i) {
        if(last.count(c[i])) {
            pev[i] = last[c[i]];
        }
        last[c[i]] = i;
        inc(pev[i] + 1);
        for(auto c : qur[i]) {
            ret[c.second.first] += get(c.first) * c.second.second;
            //cout << ret[c.second.first] << '\n';

        }
    }

    forn(i, q) cout << ret[i] << '\n';

    // 1 6 4 1 2 2 8
    // (1, 3) (2, 5)
    // 3 0 0 0 0 0 0


}
void test_case() {
    int t;
    cin >> t;
    while(t--)solve();
}
int main() {



    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    solve();





}

Subtask #1
100.0 / 100.0

# 결과 실행 시간 메모리 Grader output
1 Correct 19 ms 26860 KB Output is correct
2 Correct 19 ms 26860 KB Output is correct
3 Correct 23 ms 26860 KB Output is correct
4 Correct 19 ms 26860 KB Output is correct
5 Correct 19 ms 26860 KB Output is correct
6 Correct 20 ms 26860 KB Output is correct
7 Correct 20 ms 26860 KB Output is correct
8 Correct 33 ms 26860 KB Output is correct
9 Correct 44 ms 26860 KB Output is correct
10 Correct 58 ms 26860 KB Output is correct
11 Correct 50 ms 26860 KB Output is correct
12 Correct 140 ms 26860 KB Output is correct
13 Correct 226 ms 26880 KB Output is correct
14 Correct 72 ms 26860 KB Output is correct
15 Correct 287 ms 26860 KB Output is correct
16 Correct 39 ms 26860 KB Output is correct
17 Correct 124 ms 26860 KB Output is correct
18 Correct 697 ms 26988 KB Output is correct
19 Correct 1105 ms 26860 KB Output is correct
20 Correct 1297 ms 26992 KB Output is correct