POJ 3356 AGTC

http://poj.org/problem?id=3356

题意：

给出两个字符串，通过转换，添加，删除使两个字符串相等，求最少操作次数。

 

思路：

d[i][j]表示第一个字符串取到第i位，第二个字符串取到第j位时的最少操作次数。

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn = 1000 + 5;

int n1, n2;
char s1[maxn], s2[maxn];
int d[maxn][maxn];

int main()
{
    //freopen("D:\\txt.txt", "r", stdin);
    while (cin >> n1)
    {
        cin >> s1 + 1 >> n2 >> s2 + 1;
        memset(d, 0, sizeof(d));
        for (int i = 1; i <= n1; i++)
            d[i][0] = i;
        for (int i = 1; i <= n2; i++)
            d[0][i] = i;
        for (int i = 1; i <= n1; i++)
        for (int j = 1; j <= n2; j++)
        {
            if (s1[i] == s2[j])  d[i][j] = d[i - 1][j - 1];
            else
                d[i][j] = min(d[i - 1][j] + 1, min(d[i][j - 1] + 1, d[i - 1][j - 1] + 1));
        }
        cout << d[n1][n2] << endl;
    }
}