51Nod 1183 编辑距离 (DP)
题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1183
比较经典的问题，题意不多说了。跟最长公共子序列(LCS)类似。
dp[i][j]表示a字符串0~i-1，b字符串0~j-1最少编辑次数。
要是a[i] == b[j]的话，此时就不用改变，dp[i][j] = dp[i - 1][j - 1]
否则dp[i][j] = min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1
1 //#pragma comment(linker, "/STACK:102400000, 102400000") 2 #include <algorithm> 3 #include <iostream> 4 #include <cstdlib> 5 #include <cstring> 6 #include <cstdio> 7 #include <vector> 8 #include <cmath> 9 #include <ctime> 10 #include <list> 11 #include <set> 12 #include <map> 13 using namespace std; 14 typedef long long LL; 15 typedef pair <int, int> P; 16 const int N = 1e3 + 5; 17 int dp[N][N]; 18 char a[N], b[N]; 19 20 int main() 21 { 22 while(cin >> a >> b) { 23 int n = strlen(a), m = strlen(b); 24 //下面两个for初始化 25 for(int i = 1; i <= n; ++i) { 26 if(a[i - 1] == b[0]) 27 dp[i][1] = i - 1; 28 else 29 dp[i][1] = min(dp[i - 1][1] + 1, i); 30 } 31 for(int i = 1; i <= m; ++i) { 32 if(b[i - 1] == a[0]) 33 dp[1][i] = i - 1; 34 else 35 dp[1][i] = dp[1][i - 1] + 1; 36 } 37 for(int i = 2; i <= n; ++i) { 38 for(int j = 2; j <= m; ++j) { 39 if(a[i - 1] == b[j - 1]) { 40 dp[i][j] = dp[i - 1][j - 1]; 41 } else { 42 dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1; 43 } 44 } 45 } 46 cout << dp[n][m] << endl; 47 } 48 return 0; 49 }
