Tian Ji -- The Horse Racing

Time Limit : 2000/1000ms (Java/Other)???Memory Limit : 65536/32768K (Java/Other)

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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 0 題目鏈接（二）NYOJ364

設(shè)田忌為a，國王為b，如果兩隊(duì)所有未比賽的馬中，先對(duì)速度從大到小排序，然后：

一、如果a的最快速度大于b的最快，則直接a的最快與b的最快比賽，贏一場(chǎng)；

二、如果a的最快速度小于b的最快，則用a的最慢浪費(fèi)b的最快，輸一場(chǎng)；

三、如果a的最快速度等于b的最快，則：

1.如果a的最慢速度大于b的最慢，則直接a的最慢與b的最慢進(jìn)行比賽，贏一場(chǎng)；

2.如果b的最慢速度小于b的最慢，則用a的最慢浪費(fèi)b的最快，輸一場(chǎng)；

3.如果a的最慢速度等于b的最慢，即a與b的最慢與最快分別相等，則：先用a的最慢與b的最快比，然后找出新的最快和最慢，重復(fù)上面的比較，直到所有馬都比完。

代碼：

#include<stdio.h> #include<algorithm> using namespace std; bool cmp(int a,int b) {return a>b; } int main() {int n,i,win,ti,tj,ki,kj;int T[1005],K[1005];//T表示田忌，K表示Kingwhile(scanf("%d",&n)!=EOF&&n!=0){for(i=0;i<n;i++)scanf("%d",&T[i]);for(i=0;i<n;i++)scanf("%d",&K[i]);sort(T,T+n,cmp);sort(K,K+n,cmp); //對(duì)馬的速度從大到小排序ki=ti=0;kj=tj=n-1; //t表示田忌，k表示kingfor(i=0,win=0;i<n;i++) //開始時(shí)win=0，即不輸也不贏{if(T[ti]<K[ki]) //田忌最快的馬比國王最快的馬慢,用最慢的馬與國王最快的馬比{tj--;ki++;win--;}else if(T[ti]>K[ki]) //田忌最快的馬比國王最快的馬快，用最快的馬與國王最快的馬比{ki++;ti++;win++;}else //最快速度相同{if(T[tj]>K[kj]) //田忌最慢的馬比King最慢的馬快，用最慢的與king最慢的比{tj--;kj--;win++;}else if(T[tj]<K[kj])//田忌最慢的馬比King最慢的馬慢，用最慢的與king最快的比{tj--;ki++;win--;}else //最慢速度和最快速度都相同{if(T[tj]<K[ki]) //先用田忌的慢馬與king的快馬比，再用田忌的快馬與king的慢馬比{tj--;ki++;win--;}}}}printf("%d\n",win*200);}return 0; }

下面是一個(gè)簡單好想的代碼： #include<stdio.h> #include<algorithm> using namespace std; int t[1005], k[1005]; int main() {int n, i;while(~scanf("%d",&n) && n){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 ans = 0;int l1 = 0, r1 = n - 1, l2 = 0, r2 = n - 1;while(l1 <= r1){if(t[l1] > k[l2]){ans++;l1++;l2++;}else if(t[r1] > k[r2]){ans++;r1--;r2--;}else{if(t[l1] < k[r2]) ans--;l1++;r2--;}}printf("%d\n",ans*200);}return 0; }

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