最長公共子序列

	0	1	2	3	4	...	n
0	0	0	0	0	0	…	0
1	0
2	0
3	0
4	0
…	…
m	0

		0	1	2	3	4	5	6
		yi	B	D	C	A	B	A
0	xi	0	0	0	0	0	0	0
1	A	0	0	0	0	1	1	1
2	B	0	1	1	1	1	2	2
3	C	0	1	1	2	2	2	2
4	B	0	1	1	2	2	3	3
5	D	0	1	2	2	2	3	3
6	A	0	1	2	2	3	3	4
7	B	0	1	2	2	3	4	4

 

		0	1	2	3	4	5	6
		yi	B	D	C	A	B	A
0	xi
1	A		↑	↑	↑	↖	←	↖
2	B		↖	←	←	↑	↖	←
3	C		↑	↑	↖	←	↑	↑
4	B		↖	↑	↑	↑	↖	←
5	D		↑	↖	↑	↑	↑	↑
6	A		↑	↑	↑	↖	↑	↖
7	B		↖	↑	↑	↑	↖	↑

 

完整程序：

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include <string.h>
#include<bits/stdc++.h>
using namespace std;
//計算最長公共子序列的動態規劃算法
#define NUM 100
int c[NUM][NUM];
int b[NUM][NUM];
void LCSLength(int m,int n,const char x[],char y[])
{
int i,j;
//數組c的第0行、第0列置0
for (i =1;i<=m;i++)
c[i][0]=0;
for(i =1;i<=n;i++)
        c[0][i] = 0;
//根據遞推公式構造數組c
for(i=1;i<=m;i++)
for(j=1;j<=n;j++)
        {
            if (x[i]==y[j])
{
    c[i][j]=c[i-1][j-1]+1;
    b[i][j]=1;
    }//↖
            else if (c[i-1][j]>=c[i][j-1])
            {
                c[i][j]=c[i-1][j];    //↑
                b[i][j]=2;
            }
            else
            {
                c[i][j]=c[i][j-1];    //←
                b[i][j]=3;
            }
        }
}
void PrintLCS( const char x[], int i, int j)
{
    if(i == 0 || j == 0)
        return;
    if(b[i][j] == 1)
    {
PrintLCS(x,i-1,j-1);
        printf("%c ", x[i-1]);
    }
    else if(b[i][j] == 2)
        PrintLCS(x, i-1, j);
    else
        PrintLCS(x, i, j-1);
}
int main()
{
   // char x[NUM] = {"ABCBDAB"};
   // char y[NUM] = {"BDCABA"};
   cout<<"請輸入兩個字符串"<<endl;
   string x1,y1;
   cin>>x1>>y1;
    char x[NUM];
   char y[NUM];
   strncpy(x,x1.c_str(),x1.length()+1);///s.c_str() 字符強制轉化爲字符串函數
   	///長度必須加1，\0還佔一個位置
   	 strncpy(y,y1.c_str(),y1.length()+1);
    int m, n;
    m = strlen(x);
    n = strlen(y);
    LCSLength(m,n,x,y);
    PrintLCS(x, m, n);
    return 0;
}