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 -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

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

200
0

思路

1.先看双方目前最好的马
1.1如果田胜，则两者最强的马比赛
1.2如果王胜，则将田最弱的和王最强的比赛
1.3两者相等
1.3.1 比较两者目前最差的马
1.3.1.1 如果田胜，则两者最差的马比赛
1.3.1.2 如果田不胜，则将田最弱和王最强比赛

代码

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
bool mycompare(int a, int b)
{return a > b;
}
int main()
{int n;while (cin >> n){int ans = 0;int a[1005], b[1005];int ai = 0, bi = 0, aj = n - 1,  bj = n - 1;if(n == 0) break;for (int i = 0; i < n; i++){cin >> a[i];//输入田}sort(a, a + n, mycompare);//排序田for (int i = 0; i < n; i++){cin >> b[i];//输入王}sort(b, b + n, mycompare);//排序王while (n--){//先看双方目前最好的马if (a[ai] > b[bi])//如果田胜，则两者最强的马比赛{ans += 200;ai++;bi++;}else if (a[ai] < b[bi])//如果王胜，则将田最弱的和王最强的比赛{ans -= 200;aj--;bi++;}else//两者相等{//比较两者目前最差的马if (a[aj] > b[bj])//如果田胜，则两者最差的马比赛{ans += 200;aj--;bj--;}else//如果田不胜，则将田最弱和王最强比赛{if (a[aj] < b[bi])ans -= 200;aj--;bi++;}}}cout << ans << endl;}
}

疑惑点

有一个思路尝试实现却报wrong answer
就是先把两个序列从大到小排序
将王和田最大的标记
1.田比王大则标记田和标记王同时向后移动 ans+=200
2.田比王小则标记王向后移动，田不变
3.田和王一样大，则标记田和标记王同时向后移动
最后当王全都遍历完，ans-=没遍历到的田*200
最后的ans就是答案
试了几个案例都是正确的，一直找不到问题在哪，麻烦研究过的大佬帮忙解答，谢谢。

杭电1052-Tian Ji -- The Horse Racing 贪心算法（有思路和注释）相关推荐

杭电1042c语言循环,HDU杭电1052 Tian Ji - The Horse Racing答题报告
HDU杭电1052 Tian Ji -- The Horse Racing解题报告本人第一次写博客,希望各位大神多多指导与包涵,不足的地方还请指出,新手在此谢过啦!!! 题目描述: Time Lim ...
HDU杭电1052 Tian Ji -- The Horse Racing解题报告
本人第一次写博客,希望各位大神多多指导与包涵,不足的地方还请指出,新手在此谢过啦!!! 题目描述: Time Limit: 2000/1000 MS (Java/Others) Memory L ...
Tian Ji -- The Horse Racing 贪心算法
Problem Description Here is a famous story in Chinese history. "That was about 2300 years ago. ...
贪心法田忌赛马问题Java代码,hdoj 1052 Tian Ji - The Horse Racing【田忌赛马】【贪心】...
hdoj 1052 Tian Ji -- The Horse Racing[田忌赛马] [贪心] 思路:先按从小到大排序, 然后从最快的开始比(假设i, j 是最慢的一端, flag1, flag2是 ...
hdu 1052 Tian Ji -- The Horse Racing
Tian Ji -- The Horse Racing Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Jav ...
Tian Ji -- The Horse Racing(贪心+STL)
Problem:Tian Ji – The Horse Racing Description: Here is a famous story in Chinese history. "Tha ...
2018北大暑校acm算法训练课程 Tian Ji -- The Horse Racing 贪心
总时间限制: 5000ms 内存限制: 65536kB 描述 Here is a famous story in Chinese history. That was about 2300 years ...
HDU 1052 Tian Ji -- The Horse Racing
题目:http://acm.hdu.edu.cn/showproblem.php?pid=1052 题意:田忌赛马,贪心 #include<iostream> #include<al ...
poj hdu Tian Ji -- The Horse Racing 贪心
http://acm.hdu.edu.cn/showproblem.php?pid=1052 田忌赛马先把田忌和国王的马排序. 每次取田忌的最快的马与国王最快的马比较,有三种情况. 一,田忌最快的马 ...

杭电1052-Tian Ji -- The Horse Racing 贪心算法（有思路和注释）