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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:
Certainly, we would like to minimize the number of all possible operations.
Illustration
A G T A A G T * A G G C
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A G T * C * T G A C G CDeletion: * 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
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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 nwhere 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
题意： 把一个字符串经过最少操作步数转为另一个字符串 ——------操作可以是删除插入修改一个字符
dp[i][j]表示A[0-i] B[0-j]相等的最少步数
我们先来对B进行操作
删除的B[j] : d[i][j]=d[i][j-1]+1;
在B[j]后面插入一个： d[i][j]=d[i-1][j]+1;
删除一个数 if(B[j]==A[i]) d[i][j]=d[i-1][j-1];
else d[i][j]=d[i-1][j-1]+1;
求以上三种方法的最大d[i][j];
同理对A[i]操作也方程也是不变的
#include<iostream> #include<cstring> #include<string> using namespace std; int dp[1001][1001]; string a,b; int n,m; int Dp(int i,int j){ if(dp[i][j]==-1){ int t1=Dp(i-1,j)+1; int t2=Dp(i,j-1)+1; int t3=Dp(i-1,j-1)+(a[i-1]==b[j-1]?0:1); dp[i][j]=min(min(t1,t2),t3); } return dp[i][j]; } int main(){ while(cin>>n>>a>>m>>b){ memset(dp,-1,sizeof(dp)); int t=max(n,m); for(int i=0;i<=t;i++){ dp[0][i]=i; dp[i][0]=i; } cout<<Dp(n,m)<<endl; } return 0; }
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原文地址：http://www.cnblogs.com/demodemo/p/4732482.html
