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POJ 3356 AGTC （编辑距离 DP）

2024-07-15 23:09:54   220人阅读

Description

Let x and y be two strings over some finite alphabet A. We would like to transformx 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 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
|  |  |        |     |     |  |
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 inx is m and the number of letters in y is n wheren ≥ 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 stringx 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 stringx into a string y.

Sample Input

10 AGTCTGACGC
11 AGTAAGTAGGC

Sample Output

4


设dp[i][j]为第一个串到i为止，第二个串到j为止 的最少次数。

dp[i][j] = dp[i-1][j-1] (s[i] == s[j])

dp[i][j] = Max{dp[i][j] , dp[i-1][j]+1, dp[i][j-1]+1,  dp[i-1][j-1]+1}

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#define lson o << 1, l, m
#define rson o << 1|1, m+1, r
using namespace std;
typedef long long LL;
const int MAX=0x3f3f3f3f;
const int maxn = 1000+10;
int n, m, dp[maxn][maxn];
char a[maxn], b[maxn];
int Min(int x,int y, int z, int w){
    int ans = MAX;
    if(x < ans) ans = x;
    if(y < ans) ans = y;
    if(z < ans) ans = z;
    if(w < ans) ans = w;
    return ans;
}
int main()
{
    while(~scanf("%d",&n)) {
        scanf("%s", a+1);
        scanf("%d%s",&m, b+1);
        memset(dp, MAX,sizeof(dp));
        for(int i= 0; i<= m; i++) dp[0][i]= i;
        for(int i= 0; i<= n; i++) dp[i][0]= i;
        for(int i= 1; i<= n; i++)
            for(int j= 1; j<= m; j++){
                if(a[i]== b[j]) dp[i][j]= dp[i-1][j-1];
                dp[i][j]= Min(dp[i][j], dp[i-1][j-1]+1, dp[i-1][j]+1, dp[i][j-1]+1);
            }
        printf("%d\n", dp[n][m]);
    }
    return 0;
}




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