Divide and Conquer DP.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
#include "../template.h"

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

inline ll 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)+1) {
        ll 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+1) FOR(j,1,n+1) {
        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+1) dp[1][i] = inf;

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