POJ

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

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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

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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：
给出字符串s1。s2。
求出将s1通过下列三种操作：
1.插入一个字符
2.删除一个字符
3.改变一个字符
变换成字符s2所需要的最小操作次数。

题解：
定义dp[i][j]表示将s1的前i个和s2的前j个进行匹配所需的最小值。
可以将s1的第i个删掉，即dp[i-1][j]+1
或者将s2的第j个删掉，即dp[i][j-1]+1
或者将s1的第i个和s2的第j个进行匹配，根据它们是否相同来判断dp[i-1][j-1]是否+1
【依旧只是对拍了而已。
update：10.29 代码没问题，但是poj很坑的是多组数据。。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 1000 + 10;int n1,n2;int dp[N][N];char s1[N],s2[N];int main(){    while(~scanf("%d%s",&n1,s1+1)){        scanf("%d%s",&n2,s2+1);        memset(dp,63,sizeof(dp));        for(int i=0;i<=max(n1,n2);++i) dp[i][0]=dp[0][i]=i;        for(int i=1;i<=n1;++i){            for(int j=1;j<=n2;++j){                dp[i][j]=min(dp[i][j-1]+1,dp[i-1][j]+1);                dp[i][j]=min(dp[i][j],dp[i-1][j-1]+(s1[i]!=s2[j]));            }        }        printf("%d\n",dp[n1][n2]);    }    return 0;}