HDU 5510 Bazinga string + hash

Bazinga

Time limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 2078 Accepted Submission (s): 642

problem DescriptionLadies and gentlemen, please sit up straight.
Don ' t tilt your head. I ' m serious.

ForNGiven stringss1,s2,?,sn , labelled from1ToN, you should find the largesti (1≤i≤n) such that there exists an integerJ (1≤J<i) andSJ is not a substring ofSi .

A substring of a string si is another string, occurs in si. For example, the ' Ruiz ' is a substring of ' ruizhang ', and ' Rzhang ' are not a substring of ' Ruizhang '.  

InputThe first line contains an integerT (1≤t≤) Which is the number of the test cases.
For each test case, the first line is the positive integern (1≤n≤) and in the followingNLines list is the stringss1,s2,?,sn .
All strings is given in lower-case letters and strings is no longer than Letters. 

Outputfor each test case, the output of the largest label you get. If It does not exist, output −1.  

Sample Input45ABABCZABCABCDZABCD4YOULOVINYOUABOUTLOVINYOUALLABOUTLOVINYOU5DEDEFABCDABCDEABCDEF3ABACCC 

Sample OutputCase #1:4Case #2: -1case #3:4Case #4:3 

Source2015ACM/ICPC Asia Shenyang Station-Replay (thanks to Tohoku University)Test instructions:, in turn, give n strings, find the subscript maximum and cannot contain the string of the preceding string (that is, all the strings are the substring of the string).

#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <vector > #include <queue> #include <cstring> #include <string> #include <algorithm>using namespace Std;typedef Long long ll;typedef unsigned long long ull; #define MM (A, B) memset (A,b,sizeof (a)); #define INF 0x7f7f7f7f#def ine for (i,n) for (int i=1;i<=n;i++) #define CT continue; #define PF printf#define SC scanfconst int Mod=1000000007;const i    NT N=505+10;ull seed=13331;ull num[n];int A[n];char s[505][2005];bool Inter (int id,int k) {int len=strlen (s[id]);    Ull tmp=0,big=0,base=1;            for (int i=0;i<len;i++) {tmp=tmp*seed+s[id][i];        Big=big*seed+s[k][i];    } for (int i=0;i<len-1;i++) base*=seed;    if (Tmp==big) return true;        for (int i=len;s[k][i]!= ' n '; i++) {big= (big-s[k][i-len]*base) *seed+s[k][i];    if (big==tmp) return true; } return false;}   int main () {int cas,n,kk=0;scanf ("%d", &cas); while (cas--) {scanf ("%d", &n);        int pos=0,ans=-1;            for (int i=1;i<=n;i++) {scanf ("%s", S[i]);                while (1) {if (!pos) {a[++pos]=i;break;}                if (Inter (a[pos],i)) pos--;                     else {a[++pos]=i;                     Ans=i;                 Break    }}} printf ("Case #%d:%d\n", ++kk,ans); } return 0;}

Analysis:

1. Complexity did not think of a reasonable way down, the right way is to set a stack (or array), when the stack top string is completely contained in the current string,

Then the stack-top string is popped, and then compared, so the stack is full of strings that meet the requirements of the topic: Theoretical basis: If A is a substring of B,

So when judging, if B is completely contained in C, then a must also be completely contained in C, if B is not completely contained in C, then C must be a string that conforms to the requirement, bouncing in .

total complexity: 2*500*2000=2*10^6;2. I made a mistake in computing base because base is the l-1 of seed, so I'm going to loop the L-times and divide the seed once, and it turns out to be wrong . because the ull is 64-bit, more than 64 automatic overflow, equivalent to modulo, so base super-ull, so that the first cycle and then apart must be wrong

HDU 5510 Bazinga string + hash