首页 > ACM题库 > HDU-杭电 > HDU 1159 Common Subsequence-动态规划-[解题报告] C++
2013
12-03

HDU 1159 Common Subsequence-动态规划-[解题报告] C++

Common Subsequence

问题描述 :

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, xij = 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.
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. 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.

样例输入:

abcfbc abfcab
programming contest 
abcd mnp

样例输出:

4
2
0

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159

题目大意:给出两个字符串,求两个字符串的最长公共字串。

思路:慢慢重心开始有贪心转向动态规划了,这题就是简单的动态规划题。以题目的第一组测试数据为例。abcfbc
abfcab。

辅助空间变化示意图



可以看出:

F[i][j]=F[i-1][j-1]+1;(a[i]==b[j])

F[i][j]=max(F[i-1][j],F[i][j-1])(a[i]!=b[j]);

n由于F(i,j)只和F(i-1,j-1), F(i-1,j)和F(i,j-1)有关, 而在计算F(i,j)时, 只要选择一个合适的顺序, 就可以保证这三项都已经计算出来了, 这样就可以计算出F(i,j). 这样一直推到f(len(a),len(b))就得到所要求的解了.

代码:

#include<stdio.h>
#include<string.h>
int f[1001][1001];//**1001*1001太大不能定义在主函数,否则直接停止编译**//
int main()
{
    char a[1001],b[1001];
    int i,j,len1,len2;
    while(~scanf("%s %s",a,b))
    {
        len1=strlen(a);
        len2=strlen(b);
        for(i=0;i<=len1;i++)
        {
            f[i][0]=0;
        }
        for(i=0;i<=len2;i++)
        {
            f[0][i]=0;
        }
        for(i=1;i<=len1;i++)
        {
            for(j=1;j<=len2;j++)
            {
                if(a[i-1]==b[j-1])
                {
                    f[i][j]=f[i-1][j-1]+1;
                }
                else
                {
                    f[i][j]=f[i-1][j]>f[i][j-1]?f[i-1][j]:f[i][j-1];
                }
            }
        }
        printf("%d\n",f[len1][len2]);
    }
    return 0;
}




  1. Good task for the group. Hold it up for every yeara??s winner. This is a excellent oppotunity for a lot more enhancement. Indeed, obtaining far better and much better is constantly the crucial. Just like my pal suggests on the truth about ab muscles, he just keeps obtaining much better.

  2. 5.1处,反了;“上一个操作符的优先级比操作符ch的优先级大,或栈是空的就入栈。”如代码所述,应为“上一个操作符的优先级比操作符ch的优先级小,或栈是空的就入栈。”

  3. 思路二可以用一个长度为k的队列来实现,入队后判断下队尾元素的next指针是否为空,若为空,则出队指针即为所求。