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最新推荐文章于 2025-04-17 17:07:19 发布

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本文介绍了一种求解最长公共子序列(LCS)问题的新方法，该方法特别关注于子序列长度大于等于k的情况。通过O(n*n)预处理长度为1的LCS，并在连续子串上提取贡献，最终实现高效求解。

链接：戳这里

17: LCS

Time Limit: 1 Sec Memory Limit: 128 MB

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Description

Giving two strings consists of only lowercase letters, find the LCS(Longest Common Subsequence) whose all partition are not less than k in length.
Input

There are multiple test cases. In each test case, each of the first two lines is a string(length is less than 2100). The third line is a positive integer k. The input will end by EOF.
Output
For each test case, output the length of the two strings’ LCS.

Sample Input
abxccdef
abcxcdef
3
abccdef
abcdef
3
Sample Output
4
6
HINT

In the first test case, the answer is:

abxccdef

abcxcdef

The length is 4.

In the second test case, the answer is:

abccdef

abc def

The length is 6

题意：

给出两个串s,p和正整数k

求最长公共子序列的长度要求每次公共的子序列都必须>=k

思路：

O(n*n)求出长度为1的最长公共子序列

然后在长度>=k且连续的串上取贡献

在用一个二维数组dp[i][j]存下当前i行j列的anw

其实也就是简单的模拟最长公共子序列的过程,只是把一个元素换成了k个元素

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
char s[2200],p[2200];
int k;
int dp[2200][2200],g[2200][2200];
int main(){
    while(scanf("%s%s",s,p)!=EOF){
        scanf("%d",&k);
        mst(dp,0);
        mst(g,0);
        int n=strlen(s);
        int m=strlen(p);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(s[i-1]==p[j-1]) dp[i][j]=dp[i-1][j-1]+1;
            }
        }
        /*for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                cout<<dp[i][j]<<" ";
            }
            cout<<endl;
        }
        cout<<endl;*/
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(dp[i][j]>=k && dp[i-k][j-k]>=dp[i][j]-k){
                    g[i][j]=g[i-k][j-k]+k;
                    if(dp[i][j]>k) g[i][j]=max(g[i][j],g[i-1][j-1]+1);
                }
                else g[i][j]=max(g[i-1][j],g[i][j-1]);
              ///  cout<<g[i][j]<<" ";
            }
          ///  cout<<endl;
        }
        cout<<g[n][m]<<endl;
    }
    return 0;
}