POJ 3356 AGTC

AGTC Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12261   Accepted: 4602 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 4 Source Manila 2006

点击打开题目链接

由字符串 A 通过下列三种操作

　1、插入一个字符；

   2、删除一个字符；

　3、改变一个字符

变换到字符串 B 所需要的最少操作次数（感觉和最长公共子序列有点相似）

状态转移方程：
有三种情况可以导致我们上面设计的状态会发生转移。我们现在来看A[i] 和 B[j] ，
①、我们可以在 B[j]后面插入一个核苷酸（即一个字符）ch，ch==A[i]，这样做的话，
至少需要 dp[i - 1][j] + 1步操作，即 dp[i][j] = dp[i - 1][j] + 1。
②、我们可以删除 B[j]，这样的话，B[1...j] 变为A[1...i] 需要 dp[i][j - 1]步，
即 dp[i][j] = dp[i][j - 1] + 1。
③、我们也可以考虑修改 B[j]，使它变为A[i]，但是如果 B[j]本来就等于 A[i] 的话，
那修改其实相当于用了 0 步，如果 B[j] != A[i] 的话，那修改相当于用了 1 步。
所以 dp[i][j] = dp[i - 1][j - 1] + （A[i] == B[j] ? 0 : 1）。

决策：
决策就很简单了，从上面三种状态转移中选择一个最小值就可以了。

处理边界：
处理好边界非常重要，这里需要注意的是对dp[0][0....m],dp[0.....n][0]的初始化，
可以这样看，dp[0][i],就是说A[1...n]是一个空串，而B[1...m]是个长度为i的串，
很显然B串变为A串就是删除i个核苷酸。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int maxn = 1000 + 5;
char A[maxn], B[maxn];
int dp[maxn][maxn];
int len1, len2;

void init()
{
	memset(dp, 0, sizeof(dp));
	int Len = max(len1, len2);
	for (int i = 0; i <= Len; i++)	//处理边界 
		dp[i][0] = dp[0][i] = i;
}

int solve()
{
	for (int i = 0; i < len1; i++)
	{
		for (int j = 0; j < len2; j++)	//状态转移，选三种状态中最小的 
		{
			if (A[i] == B[j])
				dp[i + 1][j + 1] = min(dp[i][j], min(dp[i + 1][j] + 1, dp[i][j + 1] + 1));
			else 
				dp[i + 1][j + 1] = min(dp[i][j] + 1, min(dp[i + 1][j] + 1, dp[i][j + 1] + 1));
		}
	}
	return dp[len1][len2];
}

int main()
{
	
	while (~scanf("%d%s", &len1, A))
	{
		scanf("%d%s", &len2, B);
		init();
		printf("%d\n", solve());
	}
	
    return 0;
}