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Finding 3D range sum with O(N^5 complexity) using 1D & 2D range sums(Kadane's algorithm). Sample problem: UVA "Garbage Heap"
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| #include <iostream> | |
| #include <cstring> | |
| #include <algorithm> | |
| using namespace std; | |
| typedef long long int i64; | |
| void read() | |
| { | |
| int A, B, C; | |
| i64 matrix[30][30][30]; | |
| i64 maxSum = -(1LL << 60); | |
| cin >> A >> B >> C; | |
| for(int i = 0; i < A; ++i) | |
| { | |
| for(int j = 0; j < B; ++j) | |
| { | |
| for(int k = 0; k < C; ++k) | |
| { | |
| cin >> matrix[i][j][k]; | |
| } | |
| } | |
| } | |
| for(int i = 0; i < A; ++i) | |
| { | |
| i64 temp[30][30]; | |
| memset(temp, 0, sizeof temp); | |
| for(int j = i; j < A; ++j) | |
| { | |
| for(int k = 0; k < B; ++k) | |
| { | |
| for(int l = 0; l < C; ++l) | |
| { | |
| temp[k][l] += matrix[j][k][l]; | |
| } | |
| } | |
| //now we have to find the max 2D range sum in temp[B][C]; | |
| for(int k = 0; k < B; ++k) | |
| { | |
| i64 innerTemp[30], kadane2[30]; | |
| memset(innerTemp, 0, sizeof innerTemp); | |
| for(int l = k; l < B; ++l) | |
| { | |
| for(int m = 0; m < C; ++m) | |
| { | |
| innerTemp[m] += temp[l][m]; | |
| } | |
| //Applying Kadane's algorithm for 1D range sum | |
| memcpy(kadane2, innerTemp, sizeof innerTemp); | |
| for(int m = 0; m < C; ++m) | |
| { | |
| if(m > 0) kadane2[m] += kadane2[m-1]; | |
| if(kadane2[m] > maxSum) maxSum = kadane2[m]; | |
| if(kadane2[m] < 0) kadane2[m] = 0; | |
| } | |
| } | |
| } | |
| } | |
| } | |
| cout<<maxSum<<'\n'; | |
| } | |
| int main() | |
| { | |
| ios::sync_with_stdio(false); | |
| int T; | |
| cin >> T; | |
| for(int i = 0; i < T; ++i) | |
| { | |
| read(); | |
| if(i < T-1) cout<<'\n'; | |
| } | |
| return 0; | |
| } | |
| /* | |
| Test input: | |
| 2 | |
| 2 2 2 | |
| -1 2 0 -3 -2 -1 1 5 | |
| 3 3 3 | |
| 4 8 9 4 -5 -10 3 19 -99 4 -11 14 23 -32 -43 1 0 -1 -4 -2 23 -22 56 1 3 -2 -3 | |
| Expected result: | |
| 6 | |
| 78 | |
| */ |
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