poj 1458&poj 1159 最长公共子序列LCS

Common Subsequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 35240 Accepted: 13986

Description

A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,...,k, x_ij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.

Input

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output

For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.

Sample Input

abcfbc         abfcabprogramming    contest abcd           mnp

Sample Output

420

题目大意：直接求两个串的最长公共子序列。时间复杂度为O（n*m），n，m分别为两串的长度

看懂这个状态转移方程就可以了。

题目地址：Common Subsequence

AC代码：

#include<iostream>#include<cstring>#include<cstdio>using namespace std;char a[1005],b[1005];int dp[1005][1005];int main(){    int i,j;    int len1,len2;    while(~scanf("%s%s",a+1,b+1))    {        len1=strlen(a+1);        len2=strlen(b+1);        for(i=0;i<=1000;i++)        {            dp[i][0]=0;            dp[0][i]=0;        }        for(i=1;i<=len1;i++)            for(j=1;j<=len2;j++)            {                if(a[i]==b[j]) dp[i][j]=dp[i-1][j-1]+1;                else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);            }        /*for(i=1;i<=len1;i++)        {            for(j=1;j<=len2;j++)                cout<<dp[i][j]<<" ";            cout<<endl;        }*/        cout<<dp[len1][len2]<<endl;    }    return 0;}/*abcfbc  abfcabprogramming  contestabcd   mnpAb3bd db3bAABCDE BBBBB*/

Palindrome

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 49704 Accepted: 17097

Description

A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inserted into the string in order to obtain a palindrome. 

As an example, by inserting 2 characters, the string "Ab3bd" can be transformed into a palindrome ("dAb3bAd" or "Adb3bdA"). However, inserting fewer than 2 characters does not produce a palindrome. 

Input

Your program is to read from standard input. The first line contains one integer: the length of the input string N, 3 <= N <= 5000. The second line contains one string with length N. The string is formed from uppercase letters from 'A' to 'Z', lowercase letters from 'a' to 'z' and digits from '0' to '9'. Uppercase and lowercase letters are to be considered distinct.

Output

Your program is to write to standard output. The first line contains one integer, which is the desired minimal number.

Sample Input

5Ab3bd

Sample Output

2

题目大意：给你一个字符串，问你至少再加多少个字符使得这个字符串可以成为回文串。记得以前好像是王老师给我们讲过。由于是可以任意地方都可以插入。所以将所给字符串a倒置变为串b，然后求串a和串b的LCS的长度，再用串长度一减就可以得到答案。详见代码：

而且这个n是5000，5000*5000的数组当然开不下，所以需要用滚动数组来实现。

题目地址：Palindrome

AC代码：

#include<iostream>#include<cstring>#include<cstdio>using namespace std;char a[5005],b[5005];int dp[2][5005];   //使用滚动数组int main(){    int len,i,j;    while(cin>>len)    {        scanf("%s",a+1);        //strcpy(b+1,a+1);        //strrev(b+1);   //将a倒置赋值给b        for(i=1;i<=len;i++)            b[i]=a[len-i+1];        memset(dp,0,sizeof(dp));        for(i=1;i<=len;i++)        {            for(j=1;j<=len;j++)            {                if(a[i]==b[j])                    dp[i%2][j]=dp[(i-1)%2][j-1]+1;                else                    dp[i%2][j]=max(dp[(i-1)%2][j],dp[i%2][j-1]);            }        }        int res;        res=len-dp[len%2][len];        cout<<res<<endl;    }    return 0;}/*5Ab3bd*/