POJ 1458 Common Subsequence DP

Common Subsequence

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

POJ 1458

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 _{i_j} = 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;
}

POJ 1458 Common Subsequence DP

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