标签：dex div osi char std ble positive first single
You are given n strings s1,?s2,?...,?sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn?+?i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.
The first line contains single integer n (1?≤?n?≤?100) — the number of strings. The next n lines contain strings s1,?s2,?...,?sn (1?≤?|si|?≤?100), one per line. The total length of strings is not greater than 100.
The next line contains single integer m (1?≤?m?≤?100) — the number of operations. m lines follow, each of them contains two integers aiabd bi (1?≤?ai,?bi?≤?n?+?i?-?1) — the number of strings that are concatenated to form sn?+?i.
Print m lines, each should contain one integer — the answer to the question after the corresponding operation.
5
01
10
101
11111
0
3
1 2
6 5
4 4
1
2
0
On the first operation, a new string "0110" is created. For k?=?1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k?=?2 and greater there exist strings of length k that do not appear in this string (for k?=?2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length k?=?2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.
题意：给定n个01字符串，进行m次操作，每次将Sa,Sb，两个字符串合并成为新串Sn+i
求：对于每个新和成的字符串，存在数字k，满足长度为k的01字符全排列都存在于新和成的字符串的子串中，询问k的最大值
实际发现，k的范围其实很小(不知道为什么)
在k很小的情况下，暴力枚举全排列再判断就行
如果要k比较大，似乎有分治解法？
但网上很多题解都写了(1<<k),而且是int，据我所知，这个肯定会爆
然而都AC了，说明有某种结论(或者数据太水)
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 string s[501]; 7 int n,m,ans[501]; 8 int zyys(int x) 9 {int i,j,l; 10 for (i=1;i<=10;i++) 11 { 12 for (j=0;j<(1<<i);j++) 13 { 14 string myys; 15 for (l=0;l<i;l++) 16 if (j&(1<<l)) myys+=‘1‘; 17 else myys+=‘0‘; 18 if (s[x].find(myys)==-1) return i-1; 19 } 20 } 21 } 22 int main() 23 {int i,x,y; 24 cin>>n; 25 for (i=1;i<=n;i++) 26 { 27 cin>>s[i]; 28 } 29 cin>>m; 30 for (i=n+1;i<=n+m;i++) 31 { 32 scanf("%d%d",&x,&y); 33 s[i]=s[x]+s[y]; 34 if (s[i].length()>1000) 35 s[i]=s[i].substr(0,500)+s[i].substr(s[i].length()-500,500); 36 ans[i]=max(ans[x],max(ans[y],zyys(i))); 37 printf("%d\n",ans[i]); 38 } 39 }
标签：dex div osi char std ble positive first single
原文地址：http://www.cnblogs.com/Y-E-T-I/p/7672549.html
