动态规划（三）：最长公共子序列

给定一个序列X=，另一个序列Z=，若存在一个严格递增的X的下标序列对所有的1,2,3,…,k，都满足x(ik)=zk，则称Z是X的子序列。
比如Z=是X=B,C,B,D,A,B>的子序列。

求最长公共子序列。

Sample Input：
abcfbc abfcab
programming contest
abcd mnp

Sample Output：
4
2
0

方法一：朴素递归

#include
#include
#include
using namespace std;

const int Max = 1000 + 10;
char s1[Max], s2[Max];

int Fun(int n, int m){
	if(n == 0 || m == 0)
		return 0;
	else{
		if(s1[n] == s2[m])
			return Fun(n-1, m-1) + 1;
		else
			return max(Fun(n-1, m), Fun(n, m-1));
	}
}

int main(){
	while(scanf("%s%s", s1 + 1, s2 + 1)){
		int n = strlen(s1 + 1);
		int m = strlen(s2 + 1);
		int answer = Fun(n, m);
		cout << answer << endl;
	}
	system("pause");
	return 0;
}

方法二：递归策略+记忆法

#include
#include
#include
using namespace std;

const int Max = 1000 + 10;
char s1[Max], s2[Max];
int memo[Max][Max];

int Fun(int n, int m){
	if(memo[n][m] != -1)
		return memo[n][m];
	int answer;
	if(n == 0 || m == 0)
		answer = 0;
	else{
		if(s1[n] == s2[m])
			answer =  Fun(n-1, m-1) + 1;
		else
			answer = max(Fun(n-1, m), Fun(n, m-1));
	}
	memo[n][m] = answer;
	return answer;
}

int main(){
	while(scanf("%s%s", s1 + 1, s2 + 1)){
		int n = strlen(s1 + 1);
		int m = strlen(s2 + 1);
		for(int i = 0; i <= n; i++)
			for(int j = 0; j <= m; j++)
				memo[i][j] = -1;
		int answer = Fun(n, m);
		printf("%d\n", answer);
	}
	system("pause");
	return 0;
}

方法三：递推方程

#include
#include
#include
using namespace std;

const int Max = 1000 + 10;
char s1[Max], s2[Max];
int dp[Max][Max];

void Fun(int n, int m){
	for(int i = 0; i <= n; i++){
		for(int j = 0; j <= m; j++){
			if(i == 0 || j == 0)
				dp[i][j] = 0;
			else{
				if(s1[i] == s2[j])
					dp[i][j] = dp[i-1][j-1] + 1;
				else
					dp[i][j] = max(dp[i-1][j] , dp[i][j-1]);
			}
		}
	}
}

int main(){
	while(scanf("%s%s", s1 + 1, s2 + 1)){
		int n = strlen(s1 + 1);
		int m = strlen(s2 + 1);
		for(int i = 0; i <= n; i++)
			for(int j = 0; j <= m; j++)
				dp[i][j] = -1;
		Fun(n, m);
		int answer = dp[n][m];
		printf("%d\n", answer);
	}
	system("pause");
	return 0;
}