Created
September 17, 2021 05:22
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Solution to Algo DP - 2020
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#include <iostream> | |
using namespace std; | |
int main(){ | |
// IO: | |
cout << "Enter N and C:" << endl; | |
int N, C; | |
cin >> N; | |
cin >> C; | |
float **e_matrix = new float*[N]; | |
for (int i = 0; i<N; i++){ | |
e_matrix[i]=new float[C]; | |
} | |
for (int i = 0; i<N; i++){ | |
for (int j = 0; j < C; j++) | |
{ | |
cin >> e_matrix[i][j]; | |
} | |
} | |
cout << "N: " << N << " C: " << C << endl; | |
cout << "E:\n"; | |
for (int i = 0; i<N; i++){ | |
for (int j = 0; j < C; j++) | |
{ | |
cout << e_matrix[i][j] << "\t"; | |
} | |
cout << endl; | |
} | |
// SOLUTION: using tabulation | |
float **dp = new float*[N]; | |
for (int i = 0; i<N; i++){ | |
dp[i]=new float[C+1]; | |
} | |
int **comp = new int*[N]; | |
for (int i = 0; i<N; i++){ | |
comp[i]=new int[C+1]; | |
} | |
// init | |
for (int r = 1; r <= N; r++) | |
{ | |
// not enough catalyst | |
dp[r-1][0] = 0; | |
comp[r-1][0] = 0; | |
for (int c = 1; c < r; c++) | |
{ | |
dp[r-1][c] = 0; | |
} | |
if (r > 1) | |
{ | |
// just enough catalyst | |
dp[r-1][r] = dp[r-2][r-1]*e_matrix[r-1][0]; | |
comp[r-1][r] = 1; | |
} | |
else | |
// only one reaction | |
for (int c = 1; c < C+1; c++) | |
{ | |
dp[0][c] = e_matrix[0][c-1]; | |
comp[0][c] = c; | |
} | |
} | |
for (int r = 2; r <= N; r++) | |
{ | |
for (int c=r+1; c<=C; c++) | |
{ | |
// solving for dp[r-1][c] | |
float s_max = 0.0; | |
int zmax = 0; | |
// how much we assign to reaction r | |
for (int z=1; z<=c-r; z++) | |
{ | |
float solve = dp[r-2][c-z]*e_matrix[r-1][z-1]; | |
if (solve > s_max) | |
{ | |
s_max = solve; | |
zmax = z; | |
} | |
} | |
dp[r-1][c] = s_max; | |
comp[r-1][c] = zmax; | |
} | |
} | |
cout << endl << "output.txt" << endl; | |
cout << dp[N-1][C] << endl; | |
int left = C; | |
for (int r = N; r >= 1; r--) | |
{ | |
int z = comp[r-1][left]; | |
cout << "reaction " << (r) << ": " << z << endl; | |
left -= z; | |
} | |
return 0; | |
} |
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