TIANKENG’s restaurant(Ⅱ)

Time Limit: 16000/8000 MS (Java/Others)????Memory Limit: 130107/65536 K (Java/Others)

Problem Description After improving the marketing strategy, TIANKENG has made a fortune and he is going to step into the status of TuHao. Nevertheless, TIANKENG wants his restaurant to go international, so he decides to name his restaurant in English. For the lack of English skills, TIANKENG turns to CC, an English expert, to help him think of a property name. CC is a algorithm lover other than English, so he gives a long string S to TIANKENG. The string S only contains eight kinds of letters-------‘A’, ‘B’, ‘C’, ‘D’, ‘E’, ‘F’, ‘G’, ‘H’. TIANKENG wants his restaurant’s name to be out of ordinary, so the restaurant’s name is a string T which should satisfy the following conditions: The string T should be as short as possible, if there are more than one strings which have the same shortest length, you should choose the string which has the minimum lexicographic order. Could you help TIANKENG get the name as soon as possible?

Meanwhile, T is different from all the substrings of S. Could you help TIANKENG get the name as soon as possible?
Input The first line input file contains an integer T(T<=50) indicating the number of case.
In each test case:
Input a string S. the length of S is not large than 1000000.
Output For each test case:
Output the string t satisfying the condition.(T also only contains eight kinds of letters-------‘A’, ‘B’, ‘C’, ‘D’, ‘E’, ‘F’, ‘G’, ‘H’.)
Sample Input 3 ABCDEFGH AAABAACADAEAFAGAH ACAC
Sample Output AA BB B

題意：

1、找一個字符串s使得 s不是給定母串的子串

2、且s要最短

3、s在最短情況下字典序最小

解題思路：首先要肯定的一點(diǎn)就是，這個串s是不會超過8位的。8位串的話用dfs搜索肯定是沒問題的，關(guān)鍵是如何判斷字符串是否是子串，這里可以用哈希來解決。不過這里的哈希確實(shí)用的比較巧妙，我沒有想到，h[i][ans]表示長度為i，哈希值為ans的串是否存在。這樣直接傳s的哈希值即可。

#include<iostream> #include<cstdio> #include<cstring> using namespace std;const int maxn = 1000005; int flag,h[8][2496764]; char str[maxn],res[8];bool dfs(int top,int hash,int dep) {if(top == dep){for(int i = 0; i < 8; i++){if(h[top][hash * 8 + i] != flag){res[top] = i + 'A';res[top + 1] = 0;return true;}}return false;}for(int i = 0; i < 8; i++){res[top] = i + 'A';if(dfs(top+1,hash * 8 + i, dep)) return true;}return false; }int main() {int t;scanf("%d",&t);flag = 0;while(t--){flag++;scanf("%s",str);for(int i = 0; str[i]; i++){int ans = 0;for(int j = 0; j < 7 && str[i+j]; j++){ans = ans * 8 + str[i + j] - 'A'; h[j][ans] = flag;}}for(int i = 0; i < 8; i++)if(dfs(0,0,i)) break;puts(res);}return 0; }

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