鏈接：https://ac.nowcoder.com/acm/contest/148/J?&headNav=acm
來源：牛客網(wǎng)

Rikka with Nickname
時間限制：C/C++ 2秒，其他語言4秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld
題目描述
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.)
鏈接：https://ac.nowcoder.com/acm/contest/148/J?&headNav=acm
來源：牛客網(wǎng)

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.
輸入描述:
鏈接：https://ac.nowcoder.com/acm/contest/148/J?&headNav=acm
來源：牛客網(wǎng)

輸出描述:
For each testcase, output a single line with a single string, the result nickname.
鏈接：https://ac.nowcoder.com/acm/contest/148/J?&headNav=acm
來源：牛客網(wǎng)

示例1
輸入
復制
2
2
lubenwei
niubi
3
aa
ab
abb
輸出
復制
lubenweiubi
aabb

題意：

思路：

維護一個vector a[i] 數(shù)組，a[i]代表ans中字符i分別存在哪些位置。

對于每一個新嘗試加入的字符，我們?nèi)ector中二分查找盡可能靠前的位置，如果找不到就只能老老實實的加入到ans中。

能找到的話，把位置賦值為last，在 下一次查找中使用。

細節(jié)見代碼：

#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';}} }

轉(zhuǎn)載于:https://www.cnblogs.com/qieqiemin/p/11220038.html

總結

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