HDU 4323 Magic Number 编辑距离

题目描述：

Description
There are many magic numbers whose lengths are less than 10. Given some queries, each contains a single number, if the Levenshtein distance (see below) between the number in the query and a magic number is no more than a threshold, we call the magic number is the lucky number for that query. Could you find out how many luck numbers are there for each query?

Levenshtein distance (from Wikipedia
http://en.wikipedia.org/wiki/Levenshtein_distance):
In information theory and computer science, the Levenshtein distance is a string metric for measuring the amount of difference between two sequences. The term edit distance is often used to refer specifically to Levenshtein distance.
The Levenshtein distance between two strings is defined as the minimum number of edits needed to transform one string into the other, with the allowable edit operations being insertion, deletion, or substitution of a single character. It is named after Vladimir Levenshtein, who considered this distance in 1965.
For example, the Levenshtein distance between “kitten” and “sitting” is 3, since the following three edits change one into the other, and there is no way to do it with fewer than three edits:
1.kitten → sitten (substitution of ‘s’ for ‘k’)
2.sitten → sittin (substitution of ‘i’ for ‘e’)
3.sittin → sitting (insertion of ‘g’ at the end).

Input
There are several test cases. The first line contains a single number T shows that there are T cases. For each test case, there are 2 numbers in the first line: n (n <= 1500) m (m <= 1000) where n is the number of magic numbers and m is the number of queries.
In the next n lines, each line has a magic number. You can assume that each magic number is distinctive.
In the next m lines, each line has a query and a threshold. The length of each query is no more than 10 and the threshold is no more than 3.

Output
For each test case, the first line is “Case #id:”, where id is the case number. Then output m lines. For each line, there is a number shows the answer of the corresponding query.

Sample Input

1
5 2
656
67
9313
1178
38
87 1
9509 1 

Sample Output

Case #1:
1
0 

题目分析：

大意是给你n个字符串，再给你m组询问，每组询问中包含一个字符串s和一个数字x，请你求出上述n个字符串与这个字符串s的编辑距离小于等于x的字符串个数。
经典的DP编辑距离题目，在51nod上有一个编辑距离的教程，可以去看看，讲的很不错（我也是那里学的）。
传送门：
< http://www.51nod.com/tutorial/course.html#!courseId=3>
然后这道题套代码就可以AC。
我还有一篇博文也是编辑距离。

代码如下：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const double eps = 1e-4;
const int MAXN=1550;

char a[MAXN][20];
int n,m;
int T;

int DP(char *s1,char *s2)
{
    int dp[15][15];
    int len1=strlen(s1),len2=strlen(s2);
    memset(dp,0,sizeof(dp));
    for(int i=0; i<=len1; i++) dp[i][0]=i;
    for(int j=0; j<=len2; j++) dp[0][j]=j;
    for(int i=1; i<=len1; i++)
    {
        for(int j=1; j<=len2; 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] + (s1[i-1] != s2[j-1]));
        }
    }
    return (dp[len1][len2]);
}
int main()
{
    scanf("%d",&T);
    for(int t=1; t<=T; t++)
    {
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++) scanf("%s",a[i]+1);
        char s[15];
        int x;
        printf("Case #%d:\n",t);
        for(int i=1; i<=m; i++)
        {
            scanf(" %s %d",s+1,&x);
            int ans=0;
            for(int j=1; j<=n; j++)
            {
                int len1=strlen(s+1);
                int len2=strlen(a[j]+1);
                if (abs(len1-len2)>x) continue;
                if (DP(s+1,a[j]+1)<=x) ans++;
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}