POJ3356-AGTC-dp+字符串处理

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


题意：
每组数据给出两个字符串x、y，通过删除或添加或替换三种操作把两个字符串变得相同。
求最少操作次数。

思路：
可以先定义一个dp[n+20][m+20]的二维数组，要明白题目求的是什么，根据需要去写题，而且要时刻记住题目求的是什么，自己定义的东西的具体含义到底是什么。
dp[i][j]代表，从长度为i的a数组到长度为j的b数组最少变换次数。
经过分析，可以得出来一个思想——同化思想，删除和添加可以看作是一种操作，比如a串可以通过添加'c'这个字符变得和b串一样，那也就说明b数组可以通过删除操作，删除'c'变得和a
串一样，所以可以说明通过添加操作和删除操作所需操作次数是确定的，
所以题目可以转换成，通过添加或者替换操作，最少把a、b串变的相同需要多少步。

注意：
a[i]!=b[i],是逻辑表达式，只有两个返回值，满足条件返回值为1，不满足返回为0；
if(1-1),是属于算数表达式，注意区分。


碰到题目一定要去好好思考，好好思考，好好思考。

 

#include<string.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
const int N=1020;

char a[N],b[N];
int dp[N][N];

int main()
{
    int n,m;
    while(~scanf("%d%s%d%s",&n,a,&m,b))
    {
        memset(dp,0,sizeof(dp));
        for(int i=0;i<=n;i++)
            dp[i][0]=i;
        for(int i=0;i<=m;i++)
            dp[0][i]=i;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                dp[i][j]=min(dp[i-1][j]+1,dp[i][j-1]+1);
                dp[i][j]=min(dp[i][j],dp[i-1][j-1]+(a[i-1]!=b[j-1]));
            }
        }
        printf("%d\n",dp[n][m]);

    }
    return 0;
}