poj3356 字符串的最小编辑距离  dp

AGTC

Description

Let x and y be two strings over some finite alphabet A. We would like to transform x into y allowing only operations given below:

• Deletion: a letter in x is missing in y at a corresponding position.
• Insertion: a letter in y is missing in x at a corresponding position.
• Change: letters at corresponding positions are distinct

Certainly, we would like to minimize the number of all possible operations.

Illustration

A G T A A G T * A G G C
| | |       |   |   | |
A G T * C * T G A C G C

Deletion: * in the bottom line
Insertion: * in the top line
Change: when the letters at the top and bottom are distinct

This tells us that to transform x = AGTCTGACGC into y = AGTAAGTAGGC we would be required to perform 5 operations (2 changes, 2 deletions and 1 insertion). If we want to minimize the number operations, we should do it like

A  G  T  A  A  G  T  A  G  G  C
|  |  |        |     |     |  |
A  G  T  C  T  G  *  A  C  G  C

and 4 moves would be required (3 changes and 1 deletion).

In this problem we would always consider strings x and y to be fixed, such that the number of letters in x is m and the number of letters in y is n where n ≥ m.

Assign 1 as the cost of an operation performed. Otherwise, assign 0 if there is no operation performed.

Write a program that would minimize the number of possible operations to transform any string x into a string y.

Input

The input consists of the strings x and y prefixed by their respective lengths, which are within 1000.

Output

An integer representing the minimum number of possible operations to transform any string x into a string y.

Sample Input

10 AGTCTGACGC
11 AGTAAGTAGGC

Sample Output

4

求字符串的最短编辑距离，有点类似最长公共子序列。。前面的三个状态分别是添加，删除，替换。dp(i,j)=min(dp(i-1,j)+1,dp(i,j-1)+1,dp(i-1,j-1)+(s[i]!=t[j]));

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>

using namespace std;

const int maxn=1100;

char s[maxn],t[maxn];
int n,m;
int dp[maxn][maxn];

int main()
{
    while(scanf("%d%s%d%s",&n,s,&m,t)!=EOF){
        memset(dp,0,sizeof(dp));
        for(int i=0;i<=n;i++){
            for(int j=0;j<=m;j++){
                if(i==0) dp[i][j]=j;
                else if(j==0) dp[i][j]=i;
                else dp[i][j]=min(min(dp[i-1][j]+1,dp[i][j-1]+1),dp[i-1][j-1]+(s[i-1]!=t[j-1]));
            }
        }
        cout<<dp[n][m]<<endl;
    }
    return 0;
}