Created
March 14, 2012 23:25
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Parallel Programming
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#include <iostream> | |
#include <algorithm> | |
#include <string> | |
using namespace std; | |
int main(){ | |
int n; | |
cin>>n; | |
int D[n][n], I[n][n], G[n][n]; | |
const int g_e = 1; | |
const int g_i = 2; | |
char s1[n]; | |
char s2[n]; | |
cin.getline(s1, n); | |
cin.getline(s2, n); | |
// Forward Pass | |
for (int i = 0; i < n; ++i) | |
{ | |
for (int j=0; j< n; ++j) | |
{ | |
if (i == 0 && j == 0) | |
G[i][j] = 0; | |
if (i == 0 && j > 0) { | |
D[i][j] = G[i][j] + g_e; | |
G[i][j] = g_i + g_e *j; | |
} | |
else | |
D[i][j] = min(D[i - 1][j] + g_e, G[i - 1][j] + g_i); | |
if ( i > 0 && j == 0) { | |
I[i][j] = G[i][j] + g_e; | |
G[i][j] = g_i + g_e *i; | |
} | |
else | |
I[i][j] = min(I[i][j -1] + g_e, G[i][j - 1] + g_i); | |
if (i > 0 && j > 0) { | |
int s_cost = s1[i] == s2[j] ? 1 : 0; | |
int temp = min(D[i][j], I[i][j]); | |
G[i][j] = min(G[i -1][j -1] + s_cost, temp); | |
} | |
} | |
} | |
// Backward pass | |
int i = n -1; | |
int j = n - 1; | |
return 0; | |
} |
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