| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 12490 | Accepted: 3858 |
Description
That was about 2300 years ago. General Tian Ji was a
high official in the country Qi. He likes to play horse racing with the
king and others.Both of Tian and the king have three horses in different classes,
namely, regular, plus, and super. The rule is to have three rounds in a
match; each of the horses must be used in one round. The winner of a
single round takes two hundred silver dollars from the loser.Being the most powerful man in the country, the king has so nice
horses that in each class his horse is better than Tian's. As a result,
each time the king takes six hundred silver dollars from Tian.Tian Ji was not happy about that, until he met Sun Bin, one of the
most famous generals in Chinese history. Using a little trick due to
Sun, Tian Ji brought home two hundred silver dollars and such a grace in
the next match.It was a rather simple trick. Using his regular class horse race
against the super class from the king, they will certainly lose that
round. But then his plus beat the king's regular, and his super beat the
king's plus. What a simple trick. And how do you think of Tian Ji, the
high ranked official in China?
Were Tian Ji lives in nowadays, he will certainly laugh at himself.
Even more, were he sitting in the ACM contest right now, he may discover
that the horse racing problem can be simply viewed as finding the
maximum matching in a bipartite graph. Draw Tian's horses on one side,
and the king's horses on the other. Whenever one of Tian's horses can
beat one from the king, we draw an edge between them, meaning we wish to
establish this pair. Then, the problem of winning as many rounds as
possible is just to find the maximum matching in this graph. If there
are ties, the problem becomes more complicated, he needs to assign
weights 0, 1, or -1 to all the possible edges, and find a maximum
weighted perfect matching...
However, the horse racing problem is a very special case of
bipartite matching. The graph is decided by the speed of the horses -- a
vertex of higher speed always beat a vertex of lower speed. In this
case, the weighted bipartite matching algorithm is a too advanced tool
to deal with the problem.
In this problem, you are asked to write a program to solve this special case of matching problem.
Input
input consists of up to 50 test cases. Each case starts with a positive
integer n ( n<=1000) on the first line, which is the number of
horses on each side. The next n integers on the second line are the
speeds of Tian's horses. Then the next n integers on the third line are
the speeds of the king's horses. The input ends with a line that has a
single `0' after the last test case.
Output
Sample Input
3
92 83 71
95 87 74
2
20 20
20 20
2
20 19
22 18
0
Sample Output
200
0
0
Source
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<iomanip>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
#define PI 3.141592653589792128462643383279502
int main(){
//#ifdef CDZSC_June
//freopen("in.txt","r",stdin);
//#endif
//std::ios::sync_with_stdio(false);
int n;
int tian[],qi[];
while(scanf("%d",&n),n){
for(int i=;i<=n;i++)
cin>>tian[i];
for(int i=;i<=n;i++)
cin>>qi[i];
sort(tian+,tian+n+);
sort(qi+,qi+n+);
int t1=,tn=n,q1=,qn=n;
int sum=; while(t1<=tn){
if(tian[t1]>qi[q1]){
t1++;q1++;sum+=;
}
else if(tian[t1]==qi[q1]){
while(t1<=tn&&q1<=qn){
if(tian[tn]>qi[qn]){
tn--;qn--;sum+=;
}
else {
if(tian[t1]<qi[qn]) sum-=;
t1++;qn--;break;
}
}
}
else{
t1++;qn--;sum-=;
}
}
cout<<sum<<endl;
}
return ;
}