CSU-1060 Nearest Sequence
题意:给出三个字符串,求最长公共子序列的长度。
思路:DP,开个三维数组f[N][N][N],状态转移方程:f[i][j][k]=max(dp(i-1,j-1,k-1)+(a[i-1]==b[j-1]&&a[i-1]==c[k-1]),max(dp(i-1,j,k),max(dp(i,j-1,k),dp(i,j,k-1))))。
题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1060
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <cstring>
2 #include <cstdlib>
3 #include <cmath>
4 #include <string>
5 #include <algorithm>
6 #include <iostream>
7 using namespace std;
8 const int N=110;
9
10 string a,b,c;
11 int f[N][N][N];
12 int len;
13
14 int dp(int i,int j,int k){
15 if(i<0||j<0||k<0) return 0;
16 if(f[i][j][k]) return f[i][j][k];
17 return f[i][j][k]=max(dp(i-1,j-1,k-1)+(a[i-1]==b[j-1]&&a[i-1]==c[k-1]),max(dp(i-1,j,k),max(dp(i,j-1,k),dp(i,j,k-1))));
18 }
19
20 int main(){
21
22 // freopen("data.in","r",stdin);
23 // freopen("data.out","w",stdout);
24
25 while(cin>>a>>b>>c){
26 memset(f,0,sizeof(f));
27 int ans=dp(a.length(),b.length(),c.length());
28 printf("%d\n",ans-1);
29 }
30 return 0;
31 }