hdu 4681 String（dp）

最新推荐文章于 2024-07-19 14:00:00 发布

Wiking__acm

最新推荐文章于 2024-07-19 14:00:00 发布

阅读量796

点赞数

CC 4.0 BY-SA版权

分类专栏：动态规划

本文链接：https://blog.youkuaiyun.com/Wiking__acm/article/details/9984759

动态规划专栏收录该内容

57 篇文章

订阅专栏

本文介绍了一种解决最长公共子序列问题的算法，通过寻找两个字符串A和B中最长的公共子序列，同时确保第三个字符串C是该序列的连续子串。文章详细描述了算法流程，并提供了完整的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

String

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

Problem Description

Given 3 strings A, B, C, find the longest string D which satisfy the following rules:
a) D is the subsequence of A
b) D is the subsequence of B
c) C is the substring of D
Substring here means a consecutive subsequnce.
You need to output the length of D.

Input

The first line of the input contains an integer T(T = 20) which means the number of test cases.
For each test case, the first line only contains string A, the second line only contains string B, and the third only contains string C.
The length of each string will not exceed 1000, and string C should always be the subsequence of string A and string B.
All the letters in each string are in lowercase.

Output

For each test case, output Case #a: b. Here a means the number of case, and b means the length of D.

Sample Input

2
aaaaa
aaaa
aa
abcdef
acebdf
cf

Sample Output

Case #1: 4
Case #2: 3

Hint
For test one, D is "aaaa", and for test two, D is "acf".
左右分别跑一遍最长公共子序列，再枚举在A,B中C的位置，然后三段求和即可。复杂度O(n^2)。
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring>
#include<queue>
#include<stack>
using namespace std;
const int maxn = 1000 + 5;
const int INF = 2000000000;
typedef  pair<int, int> P;
typedef long long LL;

char a[maxn],b[maxn],c[maxn];
int dpl[maxn][maxn],dpr[maxn][maxn];
int n,m;
void pre(){
    n = strlen(a+1);
    m = strlen(b+1);
    memset(dpl,0,sizeof(dpl));
    memset(dpr,0,sizeof(dpr));
    for(int i = 0;i <= m;i++){
        dpl[0][i] = 0;
    }
    for(int i = 0;i <= n;i++){
        dpl[i][0] = 0;
    }
    for(int i = 1;i <= n;i++){
        for(int j = 1;j <= m;j++){
            if(a[i] == b[j]) dpl[i][j] = dpl[i-1][j-1]+1;
            else dpl[i][j] = max(dpl[i-1][j],dpl[i][j-1]);
        }
    }
    for(int i = 0;i <= m;i++){
        dpr[n+1][i] = 0;
    }
    for(int i = 0;i <= n;i++){
        dpr[i][m+1] = 0;
    }
    for(int i = n;i >= 1;i--){
        for(int j = m;j >= 1;j--){
            if(a[i] == b[j]) dpr[i][j] = dpr[i+1][j+1]+1;
            else dpr[i][j] = max(dpr[i+1][j],dpr[i][j+1]);
        }
    }
}

int la[maxn],lb[maxn],ra[maxn],rb[maxn];

int main(){
    int t,kase = 0;
    scanf("%d",&t);
    while(t--){
        kase++;
        scanf("%s",a+1);
        scanf("%s",b+1);
        scanf("%s",c+1);
        pre();
        int len = strlen(c+1);
        int p = 1,cnta = 0;
        while(p <= n){
            int fir = -1;
            int i,j;
            for(i = 1,j = p;i <= len && j <= n;){
                if(c[i] == a[j]){
                    if(fir == -1) fir = j;
                    i++;
                }
                j++;
            }
            p++;
            if(i == len+1){
               la[cnta] = fir;
               ra[cnta++] = j-1;
            }
        }
        p = 1;
        int cntb = 0;
        while(p <= m){
            int fir = -1;
            int i,j;
            for(i = 1,j = p;i <= len && j <= m;){
                if(c[i] == b[j]){
                    if(fir == -1) fir = j;
                    i++;
                }
                j++;
            }
            p++;
            if(i == len+1){
               lb[cntb] = fir;
               rb[cntb++] = j-1;
            }
        }
        int ans = 0;
        for(int i = 0;i < cnta;i++){
            for(int j = 0;j < cntb;j++){
                ans = max(ans,len+dpl[la[i]-1][lb[j]-1]+dpr[ra[i]+1][rb[j]+1]);
            }
        }
        printf("Case #%d: %d\n",kase,ans);
    }
    return 0;
}