POJ 1458(最长公共子序列 动态规划) 解题报告

/*___________________________________________POJ 1458题__________________________________________________ Common Subsequence Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 7023 Accepted: 2641 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. 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 abfcab programming contest abcd mnp Sample Output 4 2 0 ________________________________________________________________________________________________________*/ #include<stdio.h> #include<string.h> //递归地输出最长公共子序列 void LCS(int c[][300], char a[], int m, int n) { if(m==0 || n==0) return; if(c[m][n]==c[m-1][n-1]+1) ////串a的第m个元素等于串b的第n个元素 { LCS(c,a,m-1,n-1); putchar(a[m-1]); } else if(c[m-1][n]>c[m][n-1]) LCS(c,a,m-1,n); else LCS(c,a,m,n-1); } int main() { char a[300]={'/0'},b[300]={'/0'}; int c[300][300];//c[i][j]表示串a长度为i，串b长度为j时最长公共子序列的长度，c[m][n]为结果 while( scanf("%s%s",a,b)==2) { int i,j, m=strlen(a), n=strlen(b); for(i=0,j=0;j<=n;j++) c[i][j]=0; for(j=0,i=0;i<=m;i++) c[i][j]=0; for(i=1; i<=m; i++) for(j=1; j<=n; j++) { if(a[i-1]==b[j-1])//串a的第i个元素等于串b的第j个元素 c[i][j]=c[i-1][j-1]+1; else c[i][j]=c[i-1][j]>c[i][j-1] ? c[i-1][j] : c[i][j-1]; } printf("%d/n",c[m][n]); } //LCS(c,a,m,n); //putchar('/n'); return 0; } //算法课上的例题，动态规划求最长公共子序列