poj 1458 dp

2018-02-25  本文已影响0人  猴式智减法
#include <iostream>
#include <cstring>
using namespace std;

const int MAX_N = 1000;
char s1[MAX_N], s2[MAX_N];
int MaxLen[MAX_N][MAX_N];

void solve() {
     for (i = 1; i <= len1; i++) {
            for (j = 1; j <= len2; j++) {
                if (s1[i - 1] == s2[ j - 1])
                    MaxLen[i][j] = MaxLen[i - 1][j - 1] + 1;
                else
                    MaxLen[i][j] = max(MaxLen[i - 1][j], MaxLen[i][j - 1]);
            }
        }
        cout << MaxLen[len1][len2] << endl; 
    }
}

int main() {
    while (cin >> s1 >> s2) {
        int len1 = strlen(s1);
        int len2 = strlen(s2);


        int i, j;
        for (i = 1; i <= len1; i++) {
            MaxLen[i][0] = 0;
        }
        for (j = 1; j <= len2; j++) {
            MaxLen[0][j] = 0;
        }
        solve();
    return 0;
}
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