2018牛客网暑期ACM多校训练营（第十场）J Rikka with Nickname（二分，字符串）

链接：https://ac.nowcoder.com/acm/contest/148/J?&headNav=acm

来源：牛客网

Rikka with Nickname

时间限制：C/C++ 2秒，其他语言4秒

空间限制：C/C++ 262144K，其他语言524288K

64bit IO Format: %lld

题目描述

Sometimes you may want to write a sentence into your nickname like "lubenwei niubi". But how to change it into a single word? Connect them one by one like "lubenweiniubi" looks stupid.

To generate a better nickname, Rikka designs a non-trivial algorithm to merge a string sequence s1...sn into a single string. The algorithm starts with s=s1 and merges s2...sn into s one by one. The result of merging t into s is the shortest string r which satisfies s is a prefix of r and t is a subsequence of r.(If there are still multiple candidates, take the lexicographic order smallest one.)

链接：https://ac.nowcoder.com/acm/contest/148/J?&headNav=acm

来源：牛客网

For example, if we want to generate a nickname from "lubenwei niubi", we will merge "niubi" into "lubenwei", and the result is "lubenweiubi".

Now, given a sentence s1...sn with n words, Rikka wants you to calculate the resulting nickname generated by this algorithm.

输入描述:

链接：https://ac.nowcoder.com/acm/contest/148/J?&headNav=acm

来源：牛客网

输出描述:

For each testcase, output a single line with a single string, the result nickname.

链接：https://ac.nowcoder.com/acm/contest/148/J?&headNav=acm

来源：牛客网

示例1

输入

复制

2

2

lubenwei

niubi

3

aa

ab

abb

输出

复制

lubenweiubi

aabb

题意：

思路：

维护一个vector a[i] 数组，a[i]代表ans中字符i分别存在哪些位置。

对于每一个新尝试加入的字符，我们去vector中二分查找尽可能靠前的位置，如果找不到就只能老老实实的加入到ans中。

能找到的话，把位置赋值为last，在下一次查找中使用。

细节见代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

#include <iomanip>

#define ALL(x) (x).begin(), (x).end()

#define rt return

#define dll(x) scanf("%I64d",&x)

#define xll(x) printf("%I64d\n",x)

#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), '\0', 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)

#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl

using namespace std;

typedef long long ll;

ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}

ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}

ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}

inline void getInt(int* p);

const int maxn = 1000010;

const int inf = 0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

std::vector<int> v[50];

string ans, str;

int n;

int main()

{

    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);

    //freopen("D:\\common_text\code_stream\\out.txt","w",stdout);

    int t;

    gbtb;

    cin >> t;

    while (t--)

    {

        // MS0(f);

        for (char i = 'a'; i <= 'z'; ++i)

        {

            v[i - 'a'].clear();

        }

        cin >> n;

        ans = "";

        int last = -1;

        cin >> ans;

        rep(i, 0, sz(ans))

        {

            v[ans[i] - 'a'].push_back(i);

        }

        repd(i, 2, n)

        {

            cin >> str;

            int len = str.length();

            last = -1;

            repd(j, 0, len - 1)

            {

                if (sz(v[str[j] - 'a']))

                {

                    int id = lower_bound(ALL(v[str[j] - 'a']), last) - v[str[j] - 'a'].begin();

                    if (id == sz(v[str[j] - 'a']))

                    {

                        ans.push_back(str[j]);

                        v[str[j] - 'a'].push_back(sz(ans) - 1);

                        last = inf;

                    } else

                    {

                        last = v[str[j] - 'a'][id] + 1;

                    }

                } else

                {

                    ans.push_back(str[j]);

                    v[str[j] - 'a'].push_back(sz(ans) - 1);

                    last = inf;

                }

            }

        }

        cout << ans << endl;

    }

    return 0;

}

inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    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';

        }

    }

}

巴特西

2018牛客网暑期ACM多校训练营（第十场）J Rikka with Nickname（二分，字符串）

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