田忌賽馬
時間限制：3000 ms | 內存限制：65535 KB
難度：3
- 描述
- 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.
- 輸入
- The input consists of many 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.
- 輸出
- For each input case, output a line containing a single number, which is the maximum money Tian Ji will get, in silver dollars.
- 樣例輸入
-
3 92 83 71 95 87 74 2 20 20 20 20 2 20 19 22 18
- 樣例輸出
-
200 0 0
- 來源
這個貪心我是不會的，看了巨巨的題解才明白，這裏就借鑑一下他們的思想吧
題意：
田忌和齊王各有N匹馬，判斷怎樣比賽，使田忌淨勝場數最多。
我感覺這題的精髓就是，不管怎麼比賽，都要讓田忌的馬發揮最大價值。
當然，馬的第一要務是用來贏得比賽，而且要最大效益的贏，也就是要贏對方僅次於自己的馬。
當他不能完成這個任務的時候就要去輸，並拉對方最快的馬下水，給自己後面的隊友創造更大的勝利機會。
解題思路：
先都按從小到大排序，然後比較雙方最慢的馬
1.若田忌最慢的馬可以戰勝齊王最慢的馬，那麼就讓它戰勝那匹慢馬，勝利場次加1。（田忌最慢馬 > 齊王最慢馬）
2.若田忌最慢的馬不能戰勝齊王最慢的馬，那麼它更加不能戰勝其他的馬，那就讓它輸，而且輸給齊王最快馬，失敗場次加1。（田忌最慢馬 < 齊王最快馬）
3.若田忌最慢的馬與齊王最慢的馬速度相等。此時，不能簡單地認爲與它打成平手就是最好情況，相反，打平是下下策，爲什麼呢？
因爲自己後面的隊友很有可能戰勝此時對方的這匹慢馬，所以就算自己輸一場，隊友也能幫忙贏回一場，而勝一場，輸一場的收益和打平一場的收益是一樣的，而且自己輸的時候可以拉對方最快的馬下水，給己方最快的馬創造更大的勝利機會（因爲它失去了一個強勁的對手），也就是說己方最快的馬很可能因爲自己的犧牲再勝利一場，從這個角度看，還是自己故意輸掉比較好。
但是，還有一點需要注意，當自己放水前，如果己方最快的馬原本就比對方最快的馬快，然後還輸給對方最快的馬，那麼己方最快的馬的才華就浪費了，爲什麼？
很簡單，它原本就能贏，需要你放水麼？- -！換句話說，這種情況下，自己的犧牲沒有一點價值。
所以，在放水時，一定要保證己方最快馬不快於對方最快馬。滿足此條件後，讓己方最慢馬與對方最快馬去比賽（有可能平局），這樣，田忌的馬就得到了充分的利用。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
int a,b,m,n,i,j,T[10010],K[10010];
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
scanf("%d",&T[i]);
for(i=0;i<n;i++)
scanf("%d",&K[i]);
sort(T,T+n);
sort(K,K+n);
int t_slow=0,t_fast=n-1,k_slow=0,k_fast=n-1,win=0,lost=0;
while(t_slow<=t_fast)
{
if(T[t_slow]>K[k_slow])
{
win++;
t_slow++;
k_slow++;
}
else if(T[t_slow]<K[k_slow])
{
lost++;
t_slow++;
k_fast--;
}
else
{
if(T[t_fast]>K[k_fast])
{
win++;
t_fast--;
k_fast--;
}
else
{
if(T[t_slow]<K[k_fast])//這裏是有可能相等達成平局的
lost++;
t_slow++;
k_fast--;
}
}
}
printf("%d\n",(win-lost)*200);
}
return 0;
}