ACM--田忌賽馬--貪心--HDOJ 1052--Tian Ji -- The Horse Racing
來源：程序員人生 發布時間：2016-06-23 14:39:13 閱讀次數：2448次
HDOJ題目地址：傳送門
Tian Ji -- The Horse Racing
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25520 Accepted Submission(s): 7506
Problem Description
Here is a famous story in Chinese history.
"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 ⑴ 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
The 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
For each input case, output a line containing a single number, which is the maximum money Tian Ji will get, in silver dollars.
Sample Input
3
92 83 71
95 87 74
2
20 20
20 20
2
20 19
22 18
0
Sample Output
題解：貪心算法（這里便是羅列出了所有最優解得情況）：
每次取田忌的最快的馬與齊王最快的馬比較，有3種情況。
1、田忌最快的馬比齊王最快的快，那末直接用田忌最快的馬去贏齊王最快的馬。
2、田忌最快的馬比齊王最快的慢，那末用田忌最慢的馬去輸齊王最快的馬。
3、田忌最快的馬與齊王最快的馬速度1樣。
先用田忌最慢的馬與齊王最慢的馬比較。
若田忌比齊王快，直接贏掉齊王最慢的馬。
否則田忌最慢的馬再去與齊王最快的馬比較如果最快最慢的馬都1樣，則疏忽掉錢的變化。
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
int a[1005],b[1005];
int main(){
int n;
while(cin>>n&&n){
for(int i=0; i<n; i++)
cin>>a[i];
for(int i=0; i<n; i++)
cin>>b[i];
sort(a,a+n);
sort(b,b+n);
int s=0;
for(int i=0,j=0,k=n⑴,l=n⑴; i<=k;){
//如果田忌速度最慢的馬比齊王速度最慢的要快，直接贏
if(a[i]>b[j])s++,i++,j++;
//如果田忌的速度最快的馬比齊王速度最快的要快，直接贏
else if(a[k]>b[l])s++,k--,l--;
else{
//用田忌速度最慢的和齊王速度最快的馬比較，如果小于就直接輸
if(a[i]<b[l])s--;
i++,l--;
}
}
cout<<s*200<<endl;
}
return 0;
}
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