hihoCoder week 84 P1 : Lucky Substrings 【dp】

最新推荐文章于 2025-01-14 15:57:59 发布

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最新推荐文章于 2025-01-14 15:57:59 发布

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分类专栏：我就是DP

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本文探讨了如何判断一个字符串是否为幸运串，并按字典序输出所有幸运子串。通过两种方法实现：暴力枚举子串和利用动态规划预处理子串中不同字符的个数。

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P1 : Lucky Substrings

Time Limit: 10000ms

Case Time Limit: 1000ms

Memory Limit: 256MB

Description

A string s is LUCKY if and only if the number of different characters in s is a fibonacci number. Given a string consisting of only lower case letters, output all its lucky non-empty substrings in lexicographical order. Same substrings should be printed once.

Input

A string consisting no more than 100 lower case letters.

Output

Output the lucky substrings in lexicographical order, one per line. Same substrings should be printed once.

Sample Input

aabcd

Sample Output

a
aa
aab
aabc
ab
abc
b
bc
bcd
c
cd
d

定义一个串是lucky串：串中不同字符个数为斐波那契数。

题意：给你一个串，让你按字典序输出所有的lucky子串。

思路一：暴力枚举子串，直接统计。时间复杂度O(n^3)。

思路二：设置dp[i][j]表示str.(i, j)里面不同字符的个数。先预处理子串中不同字符个数(感觉麻烦的话直接用状压记录)，然后枚举子串，时间复杂度O(n^2)。

思路一：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 1000000
#define eps 1e-8
#define MAXN (10000+10)
#define MAXM (2000000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
string G[MAXN];
string str, cstr;
map<char, bool> fp;
map<string, bool> tp;
int cnt;
bool judge(int s, int t)
{
    fp.clear(); int sum = 0;
    for(int i = s; i <= t; i++)
    {
        if(!fp[str[i]]) sum++;
        fp[str[i]] = true;
    }
    G[cnt] = str.substr(s, t-s+1);
    return sum == 1 || sum == 2 || sum == 3 || sum == 5 || sum == 8 || sum == 13 || sum == 21;
}
int main()
{
    while(cin >> str)
    {
        int len = str.size(); cnt = 0; tp.clear();
        for(int i = 0; i < len; i++)
            for(int j = i; j < len; j++)
                if(judge(i, j) && !tp[G[cnt]]) {tp[G[cnt]] = true; cnt++;}
        sort(G, G+cnt);
        for(int i = 0; i < cnt; i++) cout << G[i] << endl;
    }
    return 0;
}

思路二：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 1000000
#define eps 1e-8
#define MAXN (10000+10)
#define MAXM (2000000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
string G[MAXN];
string str, cstr;
map<char, bool> fp;
map<string, bool> tp;
int cnt;
int dp[110][110];
bool State[1<<26];
bool judge(int sum){
    return sum == 1 || sum == 2 || sum == 3 || sum == 5 || sum == 8 || sum == 13 || sum == 21;
}
int main()
{
    while(cin >> str)
    {
        int len = str.size(); cnt = 0; tp.clear();
        for(int i = 0; i < len; i++)
        {
            for(int j = 0; j < 26; j++) State[1<<j] = false;
            for(int j = i; j < len; j++)
            {
                if(i == j) {dp[i][j] = 1; State[1<<(str[i]-'a')] = true;}
                else
                {
                    int val = str[j] - 'a';
                    dp[i][j] = dp[i][j-1] + (State[1<<val] ? 0 : 1);
                    State[1<<val] = true;
                }
            }
        }
        for(int i = 0; i < len; i++)
            for(int j = i; j < len; j++)
                if(judge(dp[i][j]) && !tp[str.substr(i, j-i+1)]) {G[cnt] = str.substr(i, j-i+1); tp[G[cnt]] = true; cnt++;}
        sort(G, G+cnt);
        for(int i = 0; i < cnt; i++) cout << G[i] << endl;
    }
    return 0;
}