Common Subsequence（公共子序列个数）

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %llu

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 abfcab
programming contest
abcd mnp
Sample Output
4
2
0

这道题是说的给你两个字符串，然后求这两个字符串最长的子序列。
思路：我们在假设两个字符串，abcde和abhge这两个字符串。下面我发一张图，就很清晰了。

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <cmath>#include <algorithm>#include <queue>#include <stack>#include <iomanip>#include <map>#define inf 0x3f3f3f3fusing namespace std;int dp[1005][1005]={};int main(){    char s1[1005],s2[1005];    while(scanf("%s %s",s1,s2)!=EOF)    {        int len1=strlen(s1),len2=strlen(s2);        for(int i=1;i<=len1;i++)            for(int j=1;j<=len2;j++)        {            if(s1[i-1]==s2[j-1])                dp[i][j]=dp[i-1][j-1]+1;            else                dp[i][j]=max(dp[i-1][j],dp[i][j-1]);        }        cout<<dp[len1][len2]<<endl;    }    return 0;}