pojAGTC（LCS,DP）

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題目連結：啊哈哈，點我點我
題意：給兩個字串，找出經過多少個操作可以使得兩個串相等。。思路：找出兩個串的最長公用子序列，然後用最大的串的長度減去最長公用子序列的長度得到的就是需要的運算元。。題目：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 AGTCTGACGC11 AGTAAGTAGGC

Sample Output

4

Source

Manila 2006
代碼為：

#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;}