POJ3356-AGTC（编辑距离）

AGTC

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 13042 Accepted: 4917
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

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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

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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
Source

Manila 2006

题目大意： 求两个字符串之间的编辑距离
解题思路： dp
对于串a[1...n],b[1...m]

dp[i][j]=dp[i−1][j−1]，a[i]==b[j]

dp[i][j]=min⎧⎩⎨dp[i−1][j]+wdel(a[i])dp[i][j−1]+wins(b[j])dp[i−1][j−1]+wsub(a[i],b[j])a[i]!=b[j]

边界：
dp[i][0]=∑ik=1wdel(a[k])
dp[0][j]=∑jk=1wins(b[k])

#include<iostream>#include<cstring>#include<string>#include<vector>#include<algorithm>#include<cstdio>#include<map>#include<set>#include<cmath>#include<cctype>#include<cstdlib>#include<list>#include<iomanip>using namespace std;typedef long long LL;const int MAXN=1e3+10;const int INF=0x3f3f3f3f;char a[MAXN],b[MAXN];int dp[MAXN][MAXN];int main(){    int n,m;    while(cin>>n)    {        cin>>a;        cin>>m>>b;        memset(dp,0,sizeof(dp));        for(int i=0;i<=n;i++)        {            dp[i][0]=i;        }        for(int j=0;j<=m;j++)        {            dp[0][j]=j;        }        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                if(a[i-1]==b[j-1])                    dp[i][j]=dp[i-1][j-1];                else                    dp[i][j]=min(dp[i-1][j]+1,min(dp[i][j-1]+1,dp[i-1][j-1]+1));            }        }        cout<<dp[n][m]<<endl;    }    return 0;}

空间优化：

#include<iostream>#include<cstring>#include<string>#include<vector>#include<algorithm>#include<cstdio>#include<map>#include<set>#include<cmath>#include<cctype>#include<cstdlib>#include<list>#include<iomanip>using namespace std;typedef long long LL;const int MAXN=1e3+10;const int INF=0x3f3f3f3f;char a[MAXN],b[MAXN];int dp[MAXN];int main(){    int n,m;    while(cin>>n)    {        cin>>a;        cin>>m>>b;        int tl,tmp;        memset(dp,0,sizeof(dp));        for(int j=0;j<=m;j++)        {            dp[j]=j;        }        for(int i=1;i<=n;i++)        {            tl=i-1;//top left            dp[0]=i;//left            for(int j=1;j<=m;j++)            {                tmp=dp[j];                if(a[i-1]==b[j-1])                    dp[j]=tl;                else                    dp[j]=min(dp[j]+1,min(dp[j-1]+1,tl+1));                tl=tmp;            }        }        cout<<dp[m]<<endl;    }    return 0;}