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

//#pragma comment(linker, "/STACK:102400000, 102400000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
typedef pair <int, int> P;
const int N = 1e3 + 5;
int dp[N][N];
char a[N], b[N];

int main()
{
    while(cin >> a >> b) {
        int n = strlen(a), m = strlen(b);
        //下面两个for初始化
        for(int i = 1; i <= n; ++i) {
            if(a[i - 1] == b[0])
                dp[i][1] = i - 1;
            else
                dp[i][1] = min(dp[i - 1][1] + 1, i);
        }
        for(int i = 1; i <= m; ++i) {
            if(b[i - 1] == a[0])
                dp[1][i] = i - 1;
            else
                dp[1][i] = dp[1][i - 1] + 1;
        }
        for(int i = 2; i <= n; ++i) {
            for(int j = 2; j <= m; ++j) {
                if(a[i - 1] == b[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
                }
            }
        }
        cout << dp[n][m] << endl;
    }
    return 0;
}