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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10895 | Accepted: 4188 |
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 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
求字符串的最短编辑距离，有点类似最长公共子序列。。
前面的三个状态分别是添加，删除，替换。
dp(i,j)=min(dp(i-1,j)+1,dp(i,j-1)+1,dp(i-1,j-1)+(s[i]!=t[j]));
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> using namespace std; const int maxn=1100; char s[maxn],t[maxn]; int n,m; int dp[maxn][maxn]; int main() { while(scanf("%d%s%d%s",&n,s,&m,t)!=EOF){ memset(dp,0,sizeof(dp)); for(int i=0;i<=n;i++){ for(int j=0;j<=m;j++){ if(i==0) dp[i][j]=j; else if(j==0) dp[i][j]=i; else dp[i][j]=min(min(dp[i-1][j]+1,dp[i][j-1]+1),dp[i-1][j-1]+(s[i-1]!=t[j-1])); } } cout<<dp[n][m]<<endl; } return 0; }
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原文地址：http://www.cnblogs.com/--560/p/4781182.html
