pojAGTC（LCS,DP）

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题目链接：

啊哈哈，点我点我

题意：给两个字符串，找出经过多少个操作可以使得两个串相等。。

思路：找出两个串的最长公共子序列，然后用最大的串的长度减去最长公共子序列的长度得到的就是需要的操作数。。

题目：

AGTC

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10015		Accepted: 3849

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

Source

Manila 2006

代码为：

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn=1000+10;
int dp[maxn][maxn];
char str1[maxn],str2[maxn];

int LCS(int len1,int len2)
{
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=len1;i++)
        for(int j=1;j<=len2;j++)
    {
        if(str1[i-1]==str2[j-1])
            dp[i][j]=dp[i-1][j-1]+1;
        else
            dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
    }
    return dp[len1][len2];
}


int main()
{
     int n,m;
     while(~scanf("%d%s",&n,str1))
     {
         scanf("%d%s",&m,str2);
         int len1=strlen(str1);
         int len2=strlen(str2);
         int ans=LCS(len1,len2);
         int max_ans=max(n,m);
         printf("%d\n",max_ans-ans);
     }
     return 0;
}

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pojAGTC（LCS,DP）

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原文地址：http://blog.csdn.net/u014303647/article/details/38454623