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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.
IllustrationA G T A A G T * A G G C | | | | | | | 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 | | | | | | | 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
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1010;
char strx[maxn], stry[maxn];
int lenx, leny, dp[maxn][maxn];
int main()
{
while( scanf("%d %s", &lenx, strx + 1) != EOF)
{
scanf("%d %s", &leny, stry + 1);
int maxv = max(lenx, leny);
dp[0][0] = 0;
for(int i = 1; i <= maxv; i++)
dp[0][i] = dp[i][0] = i;
for(int i = 1; i <= lenx; i++)
{
for(int j = 1; j <= leny; j++)
{
dp[i][j] = min(dp[i-1][j] + 1, dp[i][j-1] + 1);
if(strx[i] == stry[j])
dp[i][j] = min(dp[i][j], dp[i-1][j-1]);
else
dp[i][j] = min(dp[i][j], dp[i-1][j-1] + 1);
}
}
printf("%d\n", dp[lenx][leny]);
}
return 0;
}
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原文地址：http://blog.csdn.net/a1967919189/article/details/47705039
