Atcoder F

F - LCS

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $100100$ points

Problem Statement

You are given strings $\mathrm{ss}$ and $\mathrm{tt}$. Find one longest string that is a subsequence of both $\mathrm{ss}$ and $\mathrm{tt}$.

Notes

A subsequence of a string $\mathrm{xx}$ is the string obtained by removing zero or more characters from $\mathrm{xx}$ and concatenating the remaining characters without changing the order.

Constraints

• $\mathrm{ss}$ and $\mathrm{tt}$ are strings consisting of lowercase English letters.
• $\mathrm{1\le |s|,|t|\le 30001\le |s|,|t|\le 3000}$

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Input

Input is given from Standard Input in the following format:

$\mathrm{ss}$
$\mathrm{tt}$

Output

Print one longest string that is a subsequence of both $\mathrm{ss}$ and $\mathrm{tt}$. If there are multiple such strings, any of them will be accepted.

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Sample Input 1 Copy

Copy

axyb
abyxb

Sample Output 1 Copy

Copy

axb

The answer is axb or ayb; either will be accepted.

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Sample Input 2 Copy

Copy

aa
xayaz

Sample Output 2 Copy

Copy

aa

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Sample Input 3 Copy

Copy

a
z

Sample Output 3 Copy

Copy

The answer is  (an empty string).

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Sample Input 4 Copy

Copy

abracadabra
avadakedavra

Sample Output 4 Copy

Copy

aaadara


题意：给定两个字符串s和t，让你求出这两个字符串的最长公共子序列，并输出最长公共子序列。
思路：先通过DP求出LCS的DP信息，然后再根据DP信息输出对应的字符。
裸题主要看思路。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
using namespace std;
typedef long long ll;
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int dp[3050][3050];
char a[maxn];
char b[maxn];
int n,m;
int c[maxn];
int pre[maxn];
int lis[maxn];
int main()
{
    scanf("%s",a);
    scanf("%s",b);
    n=strlen(a);
    m=strlen(b);
    for(int i=n-1;i>=0;i--)
    {
        for(int j=m-1;j>=0;j--)
        {
            if(a[i]==b[j])
            {
                dp[i][j]=dp[i+1][j+1]+1;
            }else
            {
                dp[i][j]=max(dp[i+1][j],dp[i][j+1]);
            }
        }
    }
//    cout<<dp[n-1][m-1]<<endl;
    int i=0;
    int j=0;
    while(i<n&&j<m)
    {
        if(a[i]==b[j])
        {
            putchar(a[i]);
            i++;
            j++;
        }else if(dp[i][j]==dp[i+1][j])
        {
            i++;
        }else
        {
            j++;
        }
    }

    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '
');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}