LCS变形题
来源:互联网 发布:曜俊树x先导爱知h 编辑:程序博客网 时间:2024/05/20 19:32
1、http://poj.org/problem?id=3356
2、题目大意:
给定两个字符串,目的是将第一串转换成和第二串相同的字符串,可以有三种操作,1、可以有增加、删除和改变的操作,删除即如果y中有,x没有,则在同位置y中可以删去,改变是指,x中的字符可以改成跟Y中的字符一样的,求得是最少几步可以将x字符串转换成y字符串
此题类似于求最长公共子序列,状态转移方程为
if(a[i-1]==b[j-1])
dp[i][j]=minn(dp[i-1][j-1],dp[i-1][j]+1,dp[i][j-1]+1);
else
dp[i][j]=minn(dp[i-1][j-1],dp[i-1][j],dp[i][j-1])+1;
题目:
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.
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 AGTCTGACGC11 AGTAAGTAGGC
Sample Output
4
4、代码:
#include<iostream>
using namespace std;
int minn(int x,int y,int z)
{
if(x>y)
x=y;
if(x>z)
x=z;
return x;
}
int main()
{
int n,m,i,j,dp[1100][1100];
char a[1100],b[1100];
while(cin>>n)
{
for(i=1;i<=n;i++)
cin>>a[i];
cin>>m;
for(j=1;j<=m;j++)
cin>>b[j];
dp[0][0]=0;
int n1=max(n,m);
for(i=1;i<=n1;i++)
dp[i][0]=i;
for(i=1;i<=n1;i++)
dp[0][i]=i;
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
int t1,t2,t3;
t2=dp[i-1][j]+1;//插入
t3=dp[i][j-1]+1;//删除
if(a[i]==b[j])
t1=dp[i-1][j-1];//修改
else
t1=dp[i-1][j-1]+1;//修改
dp[i][j]=minn(t1,t2,t3);
}
}
cout<<dp[n][m]<<endl;
}
return 0;
}
- LCS变形题
- poj 3356 AGTC(lcs 变形题)
- POJ 3356 AGTC LCS变形题
- HZAUoj 1015: LCS (LCS变形)
- poj1080(LCS变形)
- hdu 1080(LCS变形)
- poj1080(LCS变形)
- acdream1231(LCS变形+贪心)
- hdu1503_Advanced Fruits LCS变形
- HDU1080 【LCS变形】
- POJ 1080 Human Gene Functions LCS变形题
- hdoj 1513 Palindrome (LCS 变形)
- HZAUoj 1015: LCS 【LCS变形】 + 1016: Array C 【贪心】
- PKU 2250(变形的LCS)
- poj1080 Human Gene Functions lcs 变形
- poj1080--Human Gene Functions(dp:LCS变形)
- POJ 1159-Palindrome(DP/LCS变形)
- hdu 1080 Human Gene Functions (LCS变形)
- windbg 命令输出到log文件
- iOS多线程编程指南-前言
- jQuery 基本语法
- 语音识别一些概率知识--似然估计/最大似然估计/高斯混合模型
- 开启ssh服务
- LCS变形题
- hdu 1013 Digital Roots
- CenOS Mysql-PHP开发环境配置自己看
- log4j:WARN Please initialize the log4j system properly.
- 整数二进制表示中1的个数
- 区分汇编中的变量和标号
- 用eclipse来开发C/C++
- sql2000读写数据库时死机,数据库置疑,附加报823错误。
- 单例模式