Bazinga
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2078 Accepted Submission(s): 642
Problem Description
Ladies and gentlemen, please sit up straight.
Don't tilt your head. I'm serious.
![](http://acm.hdu.edu.cn/data/images/C645-1002-1.jpg)
For n given strings S1,S2,⋯,Sn, labelled from 1 to n, you should find the largest i (1≤i≤n) such that there exists an integer j (1≤j<i) and Sj is not a substring of Si.
A substring of a string Si is another string that occurs in Si. For example, ``ruiz" is a substring of ``ruizhang", and ``rzhang" is not a substring of ``ruizhang".
Don't tilt your head. I'm serious.
![](http://acm.hdu.edu.cn/data/images/C645-1002-1.jpg)
For n given strings S1,S2,⋯,Sn, labelled from 1 to n, you should find the largest i (1≤i≤n) such that there exists an integer j (1≤j<i) and Sj is not a substring of Si.
A substring of a string Si is another string that occurs in Si. For example, ``ruiz" is a substring of ``ruizhang", and ``rzhang" is not a substring of ``ruizhang".
Input
The first line contains an integer t (1≤t≤50) which is the number of test cases.
For each test case, the first line is the positive integer n (1≤n≤500) and in the following n lines list are the strings S1,S2,⋯,Sn.
All strings are given in lower-case letters and strings are no longer than 2000 letters.
For each test case, the first line is the positive integer n (1≤n≤500) and in the following n lines list are the strings S1,S2,⋯,Sn.
All strings are given in lower-case letters and strings are no longer than 2000 letters.
Output
For each test case, output the largest label you get. If it does not exist, output −1.
Sample Input
4
5
ab
abc
zabc
abcd
zabcd
4
you
lovinyou
aboutlovinyou
allaboutlovinyou
5
de
def
abcd
abcde
abcdef
3
a
ba
ccc
Sample Output
Case #1: 4
Case #2: -1
Case #3: 4
Case #4: 3
Source
题意:,依次给出n个串, 找出下标最大且不能包含前面所有的串的串(即前面所有的串都是这个串的子串)。
#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
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
#define SC scanf
const int mod=1000000007;
const int 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]!=' ';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
",++kk,ans);
}
return 0;
}
分析:
1.复杂度没有想到合理的办法降下来,正确的做法是,设置一个栈(或数组),当栈顶字符串完全包含于当前的字符串时,
那么就弹出栈顶字符串,然后再进行比较,因此栈内保存的全是能符合题目要求的字符串:理论依据:如果a是b的子串,
那么判断时,如果b完全包含于c,那么a也必定完全包含于c,如果b不完全包含于c,那么c肯定是符合要求的字符串,弹入。
总的复杂度:2*500*2000=2*10^6;
2.在计算base时犯了个错误,因为base是seed的(l-1)次方,所以我先循环了l次,再除以一次seed,,结果这样是错的
因为ull是64位的,超过64位自动溢出,相当于取模,所以base超了ull时,这样先循环再除肯定是错的