LCS最长公共子序列

打印路径 递归是个好方法   从哪来回哪去 

#include<iostream>
#include<string>
#include<vector>
using namespace std;
string s1 =	"ABCDDEAD";
string s2 = "BBDCCDEA";
int len;

//求s1 与 s2的最长公共子序列 
int LIS(vector<vector<int> > v, vector<vector<int> >& t){
	s1.insert(0," ");
	s2.insert(0," ");
	int len1 = s1.size();
	int len2 = s2.size();
	//下标从1开始  因为计算中涉及到下标-1  避免数组越界   
	for(int i = 1; i < len1; i ++){
		for(int j = 1; j < len2; j ++){
			if(s1[i] == s2[j]){
				v[i][j] = 1 + v[i-1][j-1];
				t[i][j] = 1;
			}
				
			else if(v[i-1][j] >= v[i][j-1]){
				v[i][j] = v[i-1][j];
				t[i][j] = 2;
			}
			else{
				v[i][j] = v[i][j-1];
				t[i][j] = 3;
			}
		}
	}
	for(int i = 1; i < len1; i ++){
		for(int j = 1; j < len2; j ++)
			cout << t[i][j];
			cout << endl;
		
	}
	return v[len1-1][len2-1];
} 
void printLCS(int i,int j,vector<vector<int> > t){
	if(i == 0 || j == 0)
		return;
	if(t[i][j] == 1){
		printLCS(i-1,j-1,t);
		cout << s1[i];
	}
	//从哪里走下来的从哪里退回去 
	else if(t[i][j] == 2)
		printLCS(i-1,j,t);
	else
		printLCS(i,j-1,t);
}
int main()
{
	vector<vector<int> > v(s1.size()+1,vector<int>(s2.size()+1,0));
	vector<vector<int> > t(s1.size()+1,vector<int>(s2.size()+1,0));
	int res = LIS(v,t);
	printLCS(s1.size()-1,s2.size()-1,t);
	cout << endl << res << endl;
}

s1[i] == s2[j]    v[i][j] = 1 + v[i-1][j-1]     如果子问题的结尾相等  当前解为 1 + 两个串各退一步

s1[i] != s2[j]     v[i][j] = max(v[i-1][j],v[i][j-1]);    如果子问题的结尾不相等  当前解为两个串各退一步的较大值

其中t数组用来  记录问题的路径