原题:
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 CDeletion: * 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 分析:dp[i][j]表示第一个字符串的前i个字符要和第二个字符串的前j个字符匹配需要的最少操作次数 初始化dp[i][0]=i,若第二个字符串为0，只好把第一个字符串全删掉，所以dp[i][0]=i 同理，dp[0][j]=j 状态转移方程：若s1[i-1]==s2[j-1],则不需要操作，那么dp[i][j]=dp[i-1][j-1] 否则，我们可能有三步操作，删除，插入，变换，所以dp[i][j]=min(dp[i-1][j]+1,dp[i][j-1]+1,dp[i-1][j-1]+1) 把删除和插入看作是一个操作，自然dp[i-1][j-1]+1对应的是变换了。
代码:
#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<memory.h>
using namespace std;
int dp[1010][1010];
char s1[1010];
char s2[1010];
int main()
{
int l1,l2;
while(scanf("%d%s%d%s",&l1,s1,&l2,s2)!=EOF)
{
for(int i=0;i<=l1;i++)
for(int j=0;j<=l2;j++)
dp[i][j]=0;
for(int i=1;i<=l1;i++) dp[i][0]=i;
for(int j=1;j<=l2;j++) dp[0][j]=j;
for(int i=1;i<=l1;i++)
for(int j=1;j<=l2;j++)
{
if(s1[i-1]==s2[j-1]) dp[i][j]=dp[i-1][j-1];
else 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]+1);
}
cout<<dp[l1][l2]<<endl;
}
}