hdu 1516 String Distance and Transform Process（编辑距离+记录路径）

String Distance and Transform Process

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 486    Accepted Submission(s): 228
Special Judge

Problem Description

String Distance is a non-negative integer that measures the distance between two strings. Here we give the definition. A transform list is a list of string, where each string, except for the last one, can be changed to the string followed by adding a character,
deleting a character or replacing a character. The length of a transform list is the count of strings minus 1 (that is the count of operations to transform these two strings). The distance between two strings is the length of a transform list from one string
to the other with the minimal length. You are to write a program to calculate the distance between two strings and give the corresponding transform list.

 

Input

Input consists a sequence of string pairs, each string pair consists two lines, each string occupies one line. The length of each string will be no more than 80.

 

Output

For each string pair, you should give an integer to indicate the distance between them at the first line, and give a sequence of command to transform string 1 to string 2. Each command is a line lead by command count, then the command. A command must be

Insert pos,value

Delete pos

Replace pos,value

where pos is the position of the string and pos should be between 1 and the current length of the string (in Insert command, pos can be 1 greater than the length), and value is a character. Actually many command lists can satisfy the request, but only one of
them is required.

 

Sample Input

abcac
bcd
aaa
aabaaaa

 

Sample Output

3
1 Delete 1
2 Replace 3,d
3 Delete 4
4
1 Insert 1,a
2 Insert 2,a
3 Insert 3,b
4 Insert 7,a

 
题目大意：给一个A 序列，转变成B序列，能做的操作有三个：删除一个元素、更改一个元素、添加一个元素。

分析：构造一个二位数组，dp[i][j] 表示 长度为 i 的 A序列，变成长度为 j 的 B 序列要做的操作数。（注意 i j 可以为0 即 空序列） 

代码如下：

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

const int maxn = 85;
char A[maxn],B[maxn];
int dp[maxn][maxn],len1,len2;

void path()
{
int i,j,step;
int index=1;
i=len1; j=len2;
step=dp[len1][len2];
while(i>0 || j>0)
{
//printf("step == %d \n",step);
//printf("i,j %d %d\n",i,j);
if(i>0 && j==0)
{
printf("%d Delete %d\n",index++,i);
i--;
continue;
}
else if(i==0 && j>0)
{
printf("%d Insert 1,%c\n",index++,B[j-1]);
j--;
continue;
}
else
{
if(step == dp[i-1][j-1] && A[i-1] == B[j-1])
{
i--; j--;
}
else if(step == dp[i-1][j-1]+1)
{
printf("%d Replace %d,%c\n",index++,i,B[j-1]);
step--; i--; j--;
}
else if(step == dp[i][j-1]+1)
{
printf("%d Insert %d,%c\n",index++,i+1,B[j-1]);
step--; j--;
}
else if( step == dp[i-1][j]+1)
{
printf("%d Delete %d\n",index++,i);
step--; i--;
}
}
}
}

int main()
{
while(scanf("%s %s",A,B)!=EOF)
{
getchar();
len1 = strlen(A);
len2 = strlen(B);
memset(dp,0,sizeof(dp));
for(int i = 0; i <= len1; i++)//初始化空序列的时候需要的操作数
dp[i][0] = i;
for(int i = 0; i <= len2; i++)
dp[0][i] = i;
for(int i = 1; i <= len1; i++)
{
for(int j = 1; j <= len2; j++)
{
int tmp = min(dp[i][j-1],dp[i-1][j]) + 1;//有三种状态可以转化到当前的dp[i][j],其中第三种分情况，如果相同就省了一步操作
int d = A[i-1] == B[j-1] ? 0 : 1;
dp[i][j] = min(tmp,dp[i-1][j-1]+d);
}
}
/*for(int i=1;i<=len1;i++)
{
for(int j=1;j<=len2;j++)
printf("%d ",dp[i][j]);
printf("\n");
}*/
printf("%d\n",dp[len1][len2]);
path();
}
return 0;
}