題目鏈接：

http://poj.org/problem?id=3356

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:

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• 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

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Certainly, we would like to minimize the number of all possible operations.

Illustration

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A G T A A G T * A G G C
| | | | | | |
A G T * C * T G A C G C

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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

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A G T A A G T A G G C
| | | | | | |
A G T C T G * A C G C

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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 分析： 兩個序列中最長的序列長度減去LCS的長度 代碼如下： #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\n",t-dp[l1][l2]);}return 0; }

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轉載于:https://www.cnblogs.com/yinbiao/p/9068107.html

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