DP/optimizations/divide_conquer.cpp

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Last update: 2022-12-26 18:28:23+08:00
Link: http://codeforces.com/blog/entry/8219
Link: http://codeforces.com/contest/321/problem/E

Depends on

template.h

Code

// http://codeforces.com/blog/entry/8219
// Divide and conquer optimization:
// Original Recurrence
//   dp[i][j] = min(dp[i-1][k] + C[k][j]) for k < j
// Sufficient condition:
//   A[i][j] <= A[i][j+1]
//   where A[i][j] = smallest k that gives optimal answer
// How to use:
//   // compute i-th row of dp from L to R. optL <= A[i][L] <= A[i][R] <= optR
//   compute(i, L, R, optL, optR)
//       1. special case L == R
//       2. let M = (L + R) / 2. Calculate dp[i][M] and opt[i][M] using O(optR - optL + 1)
//       3. compute(i, L, M-1, optL, opt[i][M])
//       4. compute(i, M+1, R, opt[i][M], optR)

// Example: http://codeforces.com/contest/321/problem/E
#include "../../template.h"

const int MN = 4011;
const int inf = 1000111000;
int n, k, cost[MN][MN], dp[811][MN];

inline int getCost(int i, int j) {
    return cost[j][j] - cost[j][i-1] - cost[i-1][j] + cost[i-1][i-1];
}

void compute(int i, int L, int R, int optL, int optR) {
    if (L > R) return ;

    int mid = (L + R) >> 1, savek = optL;
    dp[i][mid] = inf;
    FOR(k,optL,min(mid-1, optR)) {
        int cur = dp[i-1][k] + getCost(k+1, mid);
        if (cur < dp[i][mid]) {
            dp[i][mid] = cur;
            savek = k;
        }
    }
    compute(i, L, mid-1, optL, savek);
    compute(i, mid+1, R, savek, optR);
}

void solve() {
    cin >> n >> k;
    FOR(i,1,n) FOR(j,1,n) {
        cin >> cost[i][j];
        cost[i][j] = cost[i-1][j] + cost[i][j-1] - cost[i-1][j-1] + cost[i][j];
    }
    dp[0][0] = 0;
    FOR(i,1,n) dp[0][i] = inf;

    FOR(i,1,k) {
        compute(i, 1, n, 0, n);
    }
    cout << dp[k][n] / 2 << endl;
}

#line 1 "DP/optimizations/divide_conquer.cpp"
// http://codeforces.com/blog/entry/8219
// Divide and conquer optimization:
// Original Recurrence
//   dp[i][j] = min(dp[i-1][k] + C[k][j]) for k < j
// Sufficient condition:
//   A[i][j] <= A[i][j+1]
//   where A[i][j] = smallest k that gives optimal answer
// How to use:
//   // compute i-th row of dp from L to R. optL <= A[i][L] <= A[i][R] <= optR
//   compute(i, L, R, optL, optR)
//       1. special case L == R
//       2. let M = (L + R) / 2. Calculate dp[i][M] and opt[i][M] using O(optR - optL + 1)
//       3. compute(i, L, M-1, optL, opt[i][M])
//       4. compute(i, M+1, R, opt[i][M], optR)

// Example: http://codeforces.com/contest/321/problem/E
#line 1 "template.h"
#include <bits/stdc++.h>
using namespace std;

#define FOR(i,a,b) for(int i=(a),_b=(b); i<=_b; i++)
#define FORD(i,a,b) for(int i=(a),_b=(b); i>=_b; i--)
#define REP(i,a) for(int i=0,_a=(a); i<_a; i++)
#define EACH(it,a) for(__typeof(a.begin()) it = a.begin(); it != a.end(); ++it)

#define DEBUG(x) { cout << #x << " = "; cout << (x) << endl; }
#define PR(a,n) { cout << #a << " = "; FOR(_,1,n) cout << a[_] << ' '; cout << endl; }
#define PR0(a,n) { cout << #a << " = "; REP(_,n) cout << a[_] << ' '; cout << endl; }

#define sqr(x) ((x) * (x))

// For printing pair, container, etc.
// Copied from https://quangloc99.github.io/2021/07/30/my-CP-debugging-template.html
template<class U, class V> ostream& operator << (ostream& out, const pair<U, V>& p) {
    return out << '(' << p.first << ", " << p.second << ')';
}

template<class Con, class = decltype(begin(declval<Con>()))>
typename enable_if<!is_same<Con, string>::value, ostream&>::type
operator << (ostream& out, const Con& con) {
    out << '{';
    for (auto beg = con.begin(), it = beg; it != con.end(); it++) {
        out << (it == beg ? "" : ", ") << *it;
    }
    return out << '}';
}
template<size_t i, class T> ostream& print_tuple_utils(ostream& out, const T& tup) {
    if constexpr(i == tuple_size<T>::value) return out << ")"; 
    else return print_tuple_utils<i + 1, T>(out << (i ? ", " : "(") << get<i>(tup), tup); 
}
template<class ...U> ostream& operator << (ostream& out, const tuple<U...>& t) {
    return print_tuple_utils<0, tuple<U...>>(out, t);
}

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long get_rand(long long r) {
    return uniform_int_distribution<long long> (0, r-1)(rng);
}

template<typename T>
vector<T> read_vector(int n) {
    vector<T> res(n);
    for (int& x : res) cin >> x;
    return res;
}

void solve();

int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    solve();
    return 0;
}
#line 18 "DP/optimizations/divide_conquer.cpp"

const int MN = 4011;
const int inf = 1000111000;
int n, k, cost[MN][MN], dp[811][MN];

inline int getCost(int i, int j) {
    return cost[j][j] - cost[j][i-1] - cost[i-1][j] + cost[i-1][i-1];
}

void compute(int i, int L, int R, int optL, int optR) {
    if (L > R) return ;

    int mid = (L + R) >> 1, savek = optL;
    dp[i][mid] = inf;
    FOR(k,optL,min(mid-1, optR)) {
        int cur = dp[i-1][k] + getCost(k+1, mid);
        if (cur < dp[i][mid]) {
            dp[i][mid] = cur;
            savek = k;
        }
    }
    compute(i, L, mid-1, optL, savek);
    compute(i, mid+1, R, savek, optR);
}

void solve() {
    cin >> n >> k;
    FOR(i,1,n) FOR(j,1,n) {
        cin >> cost[i][j];
        cost[i][j] = cost[i-1][j] + cost[i][j-1] - cost[i-1][j-1] + cost[i][j];
    }
    dp[0][0] = 0;
    FOR(i,1,n) dp[0][i] = inf;

    FOR(i,1,k) {
        compute(i, 1, n, 0, n);
    }
    cout << dp[k][n] / 2 << endl;
}