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// 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; }