POJ3356 AGTC

AGTC

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7737		Accepted: 3074

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 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 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
此题《编程之美》上有计算两个字符串的相似度，此处用动态规划解，感觉空间还可以优化。。。


//Accepted	4096K	0MS	C++	805B	2013-04-10 22:08:01
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1010;

char a[maxn], b[maxn];
int dp[maxn][maxn];
int n, m;
//dp[i][j] 表示一个串的前i个字符和第二个字符前j个字符的“距离”。

int work() {
    for(int i = 0; i <= n; i++) {//init;
        dp[i][0] = i;
        dp[0][i] = i;
    }
    for(int i = 1; i <= m; i++) {
        for(int j = 1; j <= n; j++ ) {
            if(a[i]== b[j]) {
                dp[i][j] = dp[i-1][j-1];
            }
            else {
               int t = min(dp[i-1][j], dp[i][j-1]);
               dp[i][j] = min(t, dp[i-1][j-1]) + 1;
            }
        }
    }
    return dp[m][n];
}

int main()
{
    int res = 0;
    while(scanf("%d%s%d%s", &m, a+1, &n, b+1) != EOF) {
        res = work();
        printf("%d\n", res);
    }
    return 0;
}
/*******************
TEST:
10 AGTCTGACGC
11 AGTAAGTAGGC
*******************/