Submission #1010952

#TimeUsernameProblemLanguageResultExecution timeMemory
1010952GrindMachineRaisins (IOI09_raisins)C++17
100 / 100
394 ms36440 KiB
#include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace std; using namespace __gnu_pbds; template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; typedef long long int ll; typedef long double ld; typedef pair<int,int> pii; typedef pair<ll,ll> pll; #define fastio ios_base::sync_with_stdio(false); cin.tie(NULL) #define pb push_back #define endl '\n' #define sz(a) (int)a.size() #define setbits(x) __builtin_popcountll(x) #define ff first #define ss second #define conts continue #define ceil2(x,y) ((x+y-1)/(y)) #define all(a) a.begin(), a.end() #define rall(a) a.rbegin(), a.rend() #define yes cout << "Yes" << endl #define no cout << "No" << endl #define rep(i,n) for(int i = 0; i < n; ++i) #define rep1(i,n) for(int i = 1; i <= n; ++i) #define rev(i,s,e) for(int i = s; i >= e; --i) #define trav(i,a) for(auto &i : a) template<typename T> void amin(T &a, T b) { a = min(a,b); } template<typename T> void amax(T &a, T b) { a = max(a,b); } #ifdef LOCAL #include "debug.h" #else #define debug(...) 42 #endif /* */ const int MOD = 1e9 + 7; const int N = 50 + 5; const int inf1 = int(1e9) + 5; const ll inf2 = ll(1e18) + 5; int a[N][N], p[N][N]; int dp[N][N][N][N]; int n,m; int f(int r1, int c1, int r2, int c2){ return p[r2][c2]-p[r1-1][c2]-p[r2][c1-1]+p[r1-1][c1-1]; } int go(int l1, int r1, int l2, int r2){ if(l1 > r1 or l2 > r2) return 0; int cost = f(l1,l2,r1,r2); if(l1 == r1 and l2 == r2) return 0; if(dp[l1][r1][l2][r2] != -1) return dp[l1][r1][l2][r2]; int ans = inf1; for(int i = l1; i < r1; ++i){ amin(ans,go(l1,i,l2,r2)+go(i+1,r1,l2,r2)); } for(int i = l2; i < r2; ++i){ amin(ans,go(l1,r1,l2,i)+go(l1,r1,i+1,r2)); } ans += cost; return dp[l1][r1][l2][r2] = ans; } void solve(int test_case) { cin >> n >> m; rep1(i,n) rep1(j,m) cin >> a[i][j]; rep1(i,n) rep1(j,m) p[i][j] = p[i-1][j]+p[i][j-1]-p[i-1][j-1]+a[i][j]; memset(dp,-1,sizeof dp); int ans = go(1,n,1,m); cout << ans << endl; } int main() { fastio; int t = 1; // cin >> t; rep1(i, t) { solve(i); } return 0; }
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