hdu 1800 （字符串哈希）

Flying to the Mars

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

In the year 8888, the Earth is ruled by the PPF Empire . As the population growing , PPF needs to find more land for the newborns . Finally , PPF decides to attack Kscinow who ruling the Mars . Here the problem comes! How can the soldiers reach the Mars ? PPF convokes his soldiers and asks for their suggestions . “Rush … ” one soldier answers. “Shut up ! Do I have to remind you that there isn’t any road to the Mars from here!” PPF replies. “Fly !” another answers. PPF smiles :“Clever guy ! Although we haven’t got wings , I can buy some magic broomsticks from HARRY POTTER to help you .” Now , it’s time to learn to fly on a broomstick ! we assume that one soldier has one level number indicating his degree. The soldier who has a higher level could teach the lower , that is to say the former’s level > the latter’s . But the lower can’t teach the higher. One soldier can have only one teacher at most , certainly , having no teacher is also legal. Similarly one soldier can have only one student at most while having no student is also possible. Teacher can teach his student on the same broomstick .Certainly , all the soldier must have practiced on the broomstick before they fly to the Mars! Magic broomstick is expensive !So , can you help PPF to calculate the minimum number of the broomstick needed .
For example :
There are 5 soldiers (A B C D E)with level numbers : 2 4 5 6 4;
One method :
C could teach B; B could teach A; So , A B C are eligible to study on the same broomstick.
D could teach E;So D E are eligible to study on the same broomstick;
Using this method , we need 2 broomsticks.
Another method:
D could teach A; So A D are eligible to study on the same broomstick.
C could teach B; So B C are eligible to study on the same broomstick.
E with no teacher or student are eligible to study on one broomstick.
Using the method ,we need 3 broomsticks.
……

After checking up all possible method, we found that 2 is the minimum number of broomsticks needed.

 

Input

Input file contains multiple test cases.
In a test case,the first line contains a single positive number N indicating the number of soldiers.(0<=N<=3000)
Next N lines :There is only one nonnegative integer on each line , indicating the level number for each soldier.( less than 30 digits);

 

Output

For each case, output the minimum number of broomsticks on a single line.

 

Sample Input

410203004523434

 

Sample Output

12

 

解题思路：这道题目很明显哈希。。。。可是我自己手写的哈希居然超时了。。。不理解！！

TLE:

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int size = 10009;const int seed = 31;int n,h[size],cnt;char str[31];struct node{char str[31];int next;}Hash[90001];int calc(char *s){int len = strlen(s);int sum = 0;for(int i = len-1; i >= 0; i--)sum = (sum + s[i] - '0') % size;return sum;}void addHash(char *s){int code = calc(s);int t = h[code];strcpy(Hash[cnt].str,s);Hash[cnt].next = t;h[code] = cnt++;}bool search(char *s){int code = calc(s);int t = h[code];while(t != -1){if(strcmp(Hash[t].str,s) == 0)return true;t = Hash[t].next;}return false;}int main(){while(scanf("%d",&n)!=EOF){int ans = 0;cnt = 0;memset(h,-1,sizeof(h));for(int i = 0; i <= 90000; i++){Hash[i].next = -1;memset(Hash[i].str,0,sizeof(Hash[i].str));}for(int i = 1; i <= n; i++){getchar();scanf("%s",str);if(search(str)==true)ans++;else addHash(str);}if(ans == 0)printf("1\n");else printf("%d\n",ans);}return 0;}

网上看了别人的代码，，，写的很简单，，但很奇怪的是难道这些字符串算出的哈希值不会有冲突吗？？

AC:

#include <iostream>#include<limits.h>#include<string.h>#include<string>#include<algorithm>#include<stdio.h>using namespace std;int N;int Hash[3010];char str[40];int BKDRHash(char* s){    long long seed=131;    long long hash=0;    while(*s=='0')s++;//这点很重要，因为0000345645和34564是一个数，但是如果不处理的话会对应到不同的Hash值    while(*s)    {        hash=hash*seed+(*s++);    }    return (hash & 0x7FFFFFFF);}int cmp(const void* a,const void* b){    return *(int*)b-*(int*)a;}int main(){    setbuf(stdout,NULL);    int i,ans;    while(scanf("%d",&N)!=EOF)    {        i=0,ans=1;        for(i=0;i<N;i++)        {            scanf("%s",str);            Hash[i]=BKDRHash(str);        }        qsort(Hash,N,sizeof(Hash[0]),cmp);        int temp=1;        for(i=1;i<N;i++)//ans即为 相等的数的最大值        {            if(Hash[i]==Hash[i-1])            {                temp++;                if(temp>ans)ans=temp;            }            else temp=1;        }       printf("%d\n",ans);    }    return 0;}