POJ 3356 水LCS

题目链接：

AGTC

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13855		Accepted: 5263

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 AGTCTGACGC
11 AGTAAGTAGGC

Sample Output

Source

Manila 2006

分析：

两个序列中最长的序列长度减去LCS的长度

代码如下：

#include<cstring>
#include<cstdio>
#include<string>
#include<iostream>
#include<algorithm>
#define max_v 1005
using namespace std;
char x[max_v],y[max_v];
int dp[max_v][max_v];
int l1,l2;
int main()
{
    while(~scanf("%d %s",&l1,x))
    {
        scanf("%d %s",&l2,y);
        memset(dp,0,sizeof(dp));
        for(int i=1; i<=l1; i++)
        {
            for(int j=1; j<=l2; j++)
            {
                if(x[i-1]==y[j-1])
                {
                    dp[i][j]=dp[i-1][j-1]+1;
                }
                else
                {
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }
        int t=l1;
        if(l2>l1)
            t=l2;
        printf("%d
",t-dp[l1][l2]);
    }
    return 0;
}