poj 3356 AGTC

AGTC

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8136		Accepted: 3223

DescriptionLet 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 distinctCertainly, 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 lineInsertion: * in the top lineChange: when the letters at the top and bottom are distinctThis 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.InputThe input consists of the strings x and y prefixed by their respective lengths, which are within 1000.OutputAn 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

求最短编辑距离。类似LCS的求法。

AC代码：

#include <cstring>#include <string>#include <cstdio>#include <algorithm>#include <queue>#include <cmath>#include <vector>#include <cstdlib>#include <iostream>#define max2(a,b) ((a) > (b) ? (a) : (b))#define min2(a,b) ((a) < (b) ? (a) : (b))using namespace std;int dp[1005][1005];int main(){string a,b;int len1,len2;while(cin>>len1>>a>>len2>>b){for(int i=0;i<=len1;i++)dp[i][0]=i;for(int i=0;i<=len2;i++)dp[0][i]=i;for(int i=1;i<=len1;i++)for(int j=1;j<=len2;j++){if(a[i-1]==b[j-1])dp[i][j]=min2(dp[i-1][j-1],min2(dp[i][j-1]+1,dp[i-1][j]+1));elsedp[i][j]=min2(dp[i-1][j-1]+1,min2(dp[i][j-1]+1,dp[i-1][j]+1));}cout<<dp[len1][len2]<<endl;}return 0;}

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