HDU1159——Common Subsequence

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 35297    Accepted Submission(s): 16111

Problem 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, 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. 

 

Sample Input

abcfbc abfcabprogramming contest abcd mnp

 

Sample Output

420

 

Source

Southeastern Europe 2003

 

题意：

给出两个字符串，求两个字符串的最长公共字串的长度。

解：

动态规划基础题目。

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

else dp[i][j]=(dp[i][j-1]>p[j][i-1]?dp[i][j-1]:p[j][i-1]);

#include <stdio.h>#include <string.h> int a[1000][1000];int b[1000][1000];void LCSlength(int m,int n,char x[],char y[]){    int i,j;    for(i=1;i<=m;i++) a[i][0]=0;    for(i=1;i<=n;i++) a[0][i]=0;    for(i=0;i<m;i++)    {        for(j=0;j<n;j++)        {                 if(x[i]==y[j]){                a[i+1][j+1]=a[i][j]+1;                b[i+1][j+1]=1;            }else if(a[i][j+1]>=a[i+1][j]){                a[i+1][j+1]=a[i][j+1];                b[i+1][j+1]=2;            } else{                a[i+1][j+1]=a[i+1][j];                b[i+1][j+1]=3;            }        }    }} int main(){    char x[1000];    char y[1000];    char z[10000];    int m,n;    while(gets(z))    {        memset(x,0,sizeof(x));        memset(y,0,sizeof(y));        for(m=0;m<strlen(z);m++)        {            if(z[m]!=' ')                x[m]=z[m];            else                break;        }        int i=0;        for(n=m+1;n<strlen(z);n++)        {            if(z[n]!=' ')                y[i++]=z[n];        }        LCSlength(m,i+1,x,y);        printf("%d\n",a[m][i+1]);    }    return 0; }