poj 1458 LCS(dp)全部最长公共子序列

http://poj.org/problem?id=1458

Common Subsequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 35449 Accepted: 14100

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

引进一个二维数组c[][]，用c[i][j]记录X[i]与Y[j] 的LCS 的长度，b[i][j]记录c[i][j]是通过哪一个子问题的值求得的，以决定搜索的方向。
我们是自底向上进行递推计算，那么在计算c[i,j]之前，c[i-1][j-1]，c[i-1][j]与c[i][j-1]均已计算出来。此时我们根据X[i] = Y[j]还是X[i] != Y[j]，就可以计算出c[i][j]。

问题的递归式写成：

详细结题报告：

http://blog.csdn.net/yysdsyl/article/details/4226630

代码:

/*Source CodeProblem: 1458User: 201392210Memory: 4120KTime: 32MSLanguage: C++Result: Accepted*/#include <cstdio>#include <iostream>#include <cmath>#include <string.h>#include <stdlib.h>#include <algorithm>#include <stack>#include <vector>#include <queue>#define INF 0x3f3f3f3fusing namespace std;const int maxn=1000;char x[maxn],y[maxn];int l1,l2;int dp[maxn][maxn];void lcs(){memset(dp,0,sizeof(dp));for (int i = 1; i <= l1; ++i){for (int j = 1; j <= l2; ++j){if(x[i-1]==y[j-1])dp[i][j]=dp[i-1][j-1]+1;elsedp[i][j]=max(dp[i][j-1],dp[i-1][j]);}}}int main(){freopen("input.txt","r",stdin);while(~scanf("%s%s",x,y)){l1=strlen(x);        l2=strlen(y);lcs();cout<<dp[l1][l2]<<endl;}return 0;}