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
选择一个字符串 每次增加一个字符和另一个字符串比较
画个图就比较清楚了
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w",stdout);
#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int,int> PII;
const int MAXN = 1000 + 5;
char a[MAXN],b[MAXN];
int dp[MAXN][MAXN];
int main()
{
//FIN
while(~scanf("%s%s", a, b)){
int lena = strlen(a);
int lenb = strlen(b);
memset(dp, 0, sizeof(dp));
for(int i = 1; i <= lena ; i ++){
for(int j = 1; j <= lenb; j ++){
if(a[i - 1] == b[j - 1]){
dp[i][j] = dp[i - 1][j - 1] + 1;
}
else
dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);
}
}
printf("%d
", dp[lena][lenb]);
}
return 0;
}