hdu1516（字符串距离+输出路径）

String Distance and Transform Process

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 447    Accepted Submission(s): 222
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

abcacbcdaaaaabaaaa

 

Sample Output

31 Delete 12 Replace 3,d3 Delete 441 Insert 1,a2 Insert 2,a3 Insert 3,b4 Insert 7,a

题意是说求出字符串的最短路径，并且打印出路径

路径可以用字符串路径算法求出，字符串距离算法

代码：

#include <bits/stdc++.h>using namespace std;const int maxn=1000+5;char s[maxn],t[maxn];int dp[maxn][maxn];//表示a长度为maxn时装换到长度为maxn的b所需的最短操作int n,m;void input(){}void solve(){    n=strlen(s),m=strlen(t);    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],dp[i][j-1])+1;        dp[i][j]=min(dp[i][j],dp[i-1][j-1]+(s[i-1]!=t[j-1]));    }}void output(){    printf("%d\n",dp[n][m]);    int i=strlen(s),j=strlen(t);    int step=0;    int temp;    while(i>=1||j>=1){        if(s[i-1]!=t[j-1])            temp=1;        else            temp=0;        if(dp[i][j]==(dp[i-1][j-1]+temp)&&i>=1&&j>=1){            if(temp)                printf("%d Replace %d,%c\n",++step,i,t[j-1]);            i--,j--;        }else        if(dp[i][j]==(dp[i][j-1]+1)&&j>=1){            printf("%d Insert %d %c\n",++step,i+1,t[j-1]);            j--;        }else        if(dp[i][j]==(dp[i-1][j]+1)&&i>=1){            printf("%d Delete %d\n",++step,i);            i--;        }    }}int main(){    while(~scanf("%s%s",s,t)){        solve();        output();    }    return 0;}