51Nod 1183 编辑距离
题目链接:51Nod 1183 编辑距离
题目大意:
题解:
设\(dp[i][j]\)为字符串\(A\)的前\(i\)个字符变成字符串\(B\)的前\(j\)个字符需要的最小操作数。
若字符串\(A\)的第\(i\)个字符与字符串\(B\)的第\(j\)个字符相等,则问题变成将字符串\(A\)的前\(i-1\)个字符变成字符串\(B\)的前\(j-1\)个字符;否则,执行替换、插入或删去字符操作,操作数加一。
状态转移方程为:
\[dp[i][j] = min\{dp[i - 1][j - 1] + (A[i - 1] != B[j - 1]), dp[i - 1][j] + 1, dp[i][j - 1] + 1)\}
\]
#include <iostream>
#include <string>
using namespace std;
int dp[1010][1010];
string a, b;
int main() {
cin >> a >> b;
int lena = a.length(), lenb = b.length();
if (!lena) {
cout << lenb;
} else if (!lenb) {
cout << lena;
} else {
for (int i = 0; i <= lena; ++i) {
dp[i][0] = i;
}
for (int j = 0; j <= lenb; ++j) {
dp[0][j] = j;
}
for (int i = 1; i <= lena; ++i) {
for (int j = 1; j <= lenb; ++j) {
dp[i][j] = min(dp[i - 1][j - 1] + (a[i - 1] != b[j - 1]), min(dp[i - 1][j] + 1, dp[i][j - 1] + 1));
}
}
cout << dp[lena][lenb];
}
return 0;
}