dp编辑距离 poj3356 AGTC

传送门：点击打开链接

题意：给一个字符串A，和一个字符串B，现在能对字符串A进行删除，修改，插入等操作，问至少需要多少步才能将字符串A转换成字符串B

思路：利用动态规划，总的思路和求最长公共子序列几乎是一模一样的。这里直接写一下方程..

边界dp[i][0] = dp[0][i] = i;

当x[i] == y[j]时，i和j不需要编辑,要么删除，要么插入，要么替换

dp[i][j] = min(dp[i-1][j-1], dp[i-1][j] + 1, dp[i][j - 1] + 1);

当x[i] != y[i]时, i和j需要编辑

dp[i][j] = min(dp[i-1][j-1] + 1, dp[i-1][j] + 1, dp[i][j-1] + 1);

#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck printf("fuck")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;

const int MX = 1000 + 5;

char A[MX], B[MX];
short dp[MX][MX];

int main() {
    int m, n; //FIN;
    while(~scanf("%d%s%d%s", &m, A + 1, &n, B + 1)) {
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= m; i++) dp[i][0] = i;
        for(int i = 1; i <= n; i++) dp[0][i] = i;

        for(int i = 1; i <= m; i++) {
            for(int j = 1; j <= n; j++) {
                int t = A[i] == B[j] ? 0 : 1;
                dp[i][j] = min(dp[i - 1][j - 1] + t, min(dp[i - 1][j] + 1, dp[i][j - 1] + 1));
            }
        }

        printf("%d\n", (int)dp[m][n]);
    }
    return 0;
}