UVALive 3266 田忌赛马

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

nding 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 nd 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 nd 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

rst 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.

SampleInput

3

92 83 71

95 87 74

2

20 20

20 20

2

20 19

22 18

0

SampleOutput

200

0

0

这道题走入误区 写了超久

正确的方法是

先排序

田最快的马和齐最快的马比

1如果田快， 加200, 田和齐最快的马都--

2 否则

比最慢的马

田慢 》 齐慢

+200

否则 田慢vs 齐快

-200  或者-0

#include <stdio.h>#include <stdlib.h>#define N 1010int x[N], y[N];int cmp(const void *a, const void *b){return (*(int *)b - *(int *)a);}int main(){int n;while (scanf("%d", &n) != EOF && n) {int f1 = n - 1, f2 = n - 1, s = 0, sum = 0;for (int i = 0; i < n; i++)scanf("%d", &x[i]);for (int i = 0; i < n; i++)scanf("%d", &y[i]);qsort(x, n, sizeof(x[0]), cmp);qsort(y, n, sizeof(y[0]), cmp);for (int i = 0; i <= f1; i++) {if (x[i] > y[s]) sum += 200, s++;else {if (x[f1] > y[f2]) sum += 200, f1--, f2--, i--;else {if (x[f1] < y[s]) sum -= 200, f1--, s++, i--;else f1--, s++, i--;}}}printf("%d\n", sum);}return 0;