POJ - 1458 Common Subsequence （最长公共子序列）

最新推荐文章于 2021-05-05 17:37:05 发布

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#dp #最长公共子序列

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本文深入探讨了最长公共子序列问题，通过动态规划的方法，详细解析了如何求解两个字符串的最长公共子序列的长度，并提供了完整的代码实现。

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Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 69580		Accepted: 29165

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

思路：裸的最长公共子序列，用dp[ i ][ j ] 代表字符串 a 的前 i 个字符和字符串 b 的前 j 个字符的最长公共子序列的长度。

状态转移方程：

a[ i ] == b[ j ] 时， dp[ i ][ j ] = dp[ i - 1 ][ j - 1 ] + 1

a[ i ] != b[ j ] 时， dp[ i ][ j ] = max( dp[ i - 1 ][ j ]，dp[ i ][ j - 1 ] )

//#include <bits/stdc++.h>
#include <map>
//#include <tr1/unordered_map>
#include<limits>
#include <float.h>
#include <list>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
inline int lowbit(int x){ return x & (-x); }
inline int read(){int X = 0, w = 0; char ch = 0;while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();return w ? -X : X;}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define db double
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 10;
const int maxp = 1e3 + 10;
const LL INF = 0x3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int mod = 20090717;

char a[1010], b[1010];
int dp[1010][1010];

int main(){
    while (~scanf("%s%s", a, b)){
        int n = strlen(a), m = strlen(b);
        dp[0][0] = 0;
        For (i, 1, n)
        For (j, 1, m)
            if (a[i-1] == b[j-1]) //之所以这样是因为字符串的下标是从 0 开始的
                dp[i][j] = dp[i-1][j-1] + 1;
            else
                dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
        Pri(dp[n][m]);
    }
}