杭电1800-Flying to the Mars
2013-07-27 20:42
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Flying to the Mars
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8319 Accepted Submission(s): 2692
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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);
[align=left]Output[/align]
For each case, output the minimum number of broomsticks on a single line.
[align=left]Sample Input[/align]
4
10
20
30
04
5
2
3
4
3
4
[align=left]Sample Output[/align]
1
2题意:就要求一群士兵怎样用最少的魔法扫把飞去火星,而且能力高只能带能力低的而且只能带一个题目第一行给出n(士兵的数量)下面n行给出士兵的能力,问你需要最少需要多少扫把AC代码:
#include<iostream> #include<cstring> #include<string> #include<cstdio> #include<algorithm> const int MAX=3001; int s[MAX]; int t[MAX][MAX]; int k[MAX]; using namespace std; bool cmp(int x,int y) { return x>y; } int main() { int n,i,j,a,b,p; while(cin>>n) { for(i=0;i<n;i++) { scanf("%d",&s[i]); } sort(s,s+n,cmp); t[0][0]=s[0]; a=1; b=1; for(i=1;i<n;i++) { p=0; if(s[i]<t[0][b-1]) { t[0][b]=s[i]; b+=1; } else { for(j=1;j<a;j++) { if(s[i]<t[j][k[j]-1]) { t[j][k[j]]=s[i]; k[j]+=1; p=1; break; } } if(p!=1) { k[a]=0; t[a][k[a]]=s[i]; k[a]+=1; a+=1; } } } cout<<a<<endl; } return 0; }//此题和我写1257是同样题型,方法是一模一样,这里没写解释,所以不懂的可以看看我写的1257的解题报告
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