ACM_Notebook_new

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:warning: DP/optimizations/knuth.cpp

Depends on

Code

// http://codeforces.com/blog/entry/8219
// Original Recurrence:
//   dp[i][j] = min(dp[i][k] + dp[k][j]) + C[i][j]   for k = i+1..j-1
// Necessary & Sufficient Conditions:
//   A[i][j-1] <= A[i][j] <= A[i+1][j]
//   with A[i][j] = smallest k that gives optimal answer
// Also applicable if the following conditions are met:
//   1. C[a][c] + C[b][d] <= C[a][d] + C[b][c] (quadrangle inequality)
//   2. C[b][c] <= C[a][d]                     (monotonicity)
//   for all a <= b <= c <= d
// To use:
//   Calculate dp[i][i] and A[i][i]
//
//   FOR(len = 1..n-1)
//     FOR(i = 1..n-len) {
//       j = i + len
//       FOR(k = A[i][j-1]..A[i+1][j])
//         update(dp[i][j])
//     }
// 
// There is another type of Knuth in https://oj.vnoi.info/problem/icpc22_mn_c
// - f[i][j] = min(f[i-1][last] + cost[last+1][j])
// - cost satisfies quandrangle inequality
//   FOR(i, 1, k)
//     FORD(j, n, 1)
//       FOR(last, opt[i-1][j], opt[i][j+1])
//         update f[i][j] and A[i][j] using f[i-1][last] + cost[last+1][j]

// OPTCUT
#include "../../template.h"

const int MN = 2011;
int a[MN], dp[MN][MN], C[MN][MN], A[MN][MN];
int n;

void solve() {
    cin >> n; FOR(i,1,n) { cin >> a[i]; a[i] += a[i-1]; }
    FOR(i,1,n) FOR(j,i,n) C[i][j] = a[j] - a[i-1];

    FOR(i,1,n) dp[i][i] = 0, A[i][i] = i;

    FOR(len,1,n-1)
        FOR(i,1,n-len) {
            int j = i + len;
            dp[i][j] = 2000111000;
            FOR(k,A[i][j-1],A[i+1][j]) {
                int cur = dp[i][k-1] + dp[k][j] + C[i][j];
                if (cur < dp[i][j]) {
                    dp[i][j] = cur;
                    A[i][j] = k;
                }
            }
        }
    cout << dp[1][n] << endl;
}
#line 1 "DP/optimizations/knuth.cpp"
// http://codeforces.com/blog/entry/8219
// Original Recurrence:
//   dp[i][j] = min(dp[i][k] + dp[k][j]) + C[i][j]   for k = i+1..j-1
// Necessary & Sufficient Conditions:
//   A[i][j-1] <= A[i][j] <= A[i+1][j]
//   with A[i][j] = smallest k that gives optimal answer
// Also applicable if the following conditions are met:
//   1. C[a][c] + C[b][d] <= C[a][d] + C[b][c] (quadrangle inequality)
//   2. C[b][c] <= C[a][d]                     (monotonicity)
//   for all a <= b <= c <= d
// To use:
//   Calculate dp[i][i] and A[i][i]
//
//   FOR(len = 1..n-1)
//     FOR(i = 1..n-len) {
//       j = i + len
//       FOR(k = A[i][j-1]..A[i+1][j])
//         update(dp[i][j])
//     }
// 
// There is another type of Knuth in https://oj.vnoi.info/problem/icpc22_mn_c
// - f[i][j] = min(f[i-1][last] + cost[last+1][j])
// - cost satisfies quandrangle inequality
//   FOR(i, 1, k)
//     FORD(j, n, 1)
//       FOR(last, opt[i-1][j], opt[i][j+1])
//         update f[i][j] and A[i][j] using f[i-1][last] + cost[last+1][j]

// OPTCUT
#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 31 "DP/optimizations/knuth.cpp"

const int MN = 2011;
int a[MN], dp[MN][MN], C[MN][MN], A[MN][MN];
int n;

void solve() {
    cin >> n; FOR(i,1,n) { cin >> a[i]; a[i] += a[i-1]; }
    FOR(i,1,n) FOR(j,i,n) C[i][j] = a[j] - a[i-1];

    FOR(i,1,n) dp[i][i] = 0, A[i][i] = i;

    FOR(len,1,n-1)
        FOR(i,1,n-len) {
            int j = i + len;
            dp[i][j] = 2000111000;
            FOR(k,A[i][j-1],A[i+1][j]) {
                int cur = dp[i][k-1] + dp[k][j] + C[i][j];
                if (cur < dp[i][j]) {
                    dp[i][j] = cur;
                    A[i][j] = k;
                }
            }
        }
    cout << dp[1][n] << endl;
}
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