1408121541-hd-Tian Ji -- The Horse Racing.cpp

                                        Problem F

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

                                                                           Total Submission(s) : 6   Accepted Submission(s) : 3

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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

 

 

题目大意

       田忌赛马。

 

错误原因

       我做了双重循环，想要找出田忌可以赢的最大次数，但是却忽略了一个问题。

错误代码

<span style="font-size:18px;">#include<stdio.h>
#include<algorithm>
using namespace std;
int a[1100],b[1100];
bool cmp(int a,int b)
{
    return a>b;
}
int main()
{
    int n;
    int i,j,k,sum;
    while(scanf("%d",&n),n)
    {
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(i=0;i<n;i++)
            scanf("%d",&b[i]);
        sort(a,a+n,cmp);
        sort(b,b+n,cmp);
        sum=0;
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
            {
                if(a[i]>b[j]&&b[j]!=-1)
                {
                    sum++;
                    b[j]=-1;
                    a[i]=-1;
                    break;
                }
            }
        k=0;
        for(i=0;i<n&&a[i]!=-1;i++)
            for(j=0;j<n&&b[j]!=-1;j++)
                if(a[i]==b[j])
                {
                    a[i]=-1;
                    b[i]=-1;
                    k++;
                    break;
                }
        当数据为 2
                 96 80
                 96 70  的时候，输出结果是0，但是很显然正确答案应该是200
        我的代码会去先找可以赢的，但是却忽略了大局。这个思路行不通，
        可能模拟田忌赛马的思路走了 
        printf("%d-%d-%d ",sum,k,n-k-sum);
        printf("%d\n",200*(2*sum-n+k));
    }
    return 0;
}</span>

解题思路

       这是一道贪心题，田忌的思路就是贪心，我们可以模拟一下田忌的思路就OK了。千万要注意考虑到每种情况。

      先把a[] b[[]数组排序（我这里用的是从大到小）

1、如果田忌最快的马比齐王最快的马快，则比之

2、如果田忌最快的马比齐王最快的马慢，则用田最慢的马跟齐最快的马比  //这是贪心的第一步

3、如果田忌最快的马的速度与齐威王最快的马速度相等

3.1、如果田忌最慢的比齐威王最慢的快，则比之                         //这是贪心的第二步

3.2、如果田忌最慢的比齐威王最慢的慢，田忌慢VS齐王快

3.3、田忌最慢的与齐威王最慢的相等，田忌慢VS齐王快

正确代码

#include<stdio.h>
#include<algorithm>
using namespace std;
int a[1100],b[1100];
bool cmp(int a,int b)
{
    return a>b;
}
int main()
{
    int n;
    int i,j,k,sum;
    while(scanf("%d",&n),n)
    {
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(i=0;i<n;i++)
            scanf("%d",&b[i]);
        sort(a,a+n,cmp);
        sort(b,b+n,cmp);
        sum=0;
        k=n;
        for(i=0,j=0;i<k;)
        {
            if(a[i]>b[j])         //如果田忌最快的马比齐王最快的马快，则比之
              {
                sum++;
                i++;
                j++;
            }
            else if(a[i]<b[j])    //如果田忌最快的马比齐王最快的马慢，则用田最慢的马跟齐最快的马比  
                                  //这是贪心的第一步
              {
                k--;
                j++;
                sum--;
            }
            else if(a[i]==b[j])   //如果田忌最快的马的速度与齐威王最快的马速度相等
              {
                if(a[k-1]>b[n-1])  //如果田忌最慢的比齐威王最慢的快，则比之 
                                  //这是贪心的第二步
                   {
                    sum++;
                    n--;
                    k--;
                }
                else 
                {
                    if(a[k-1]<b[j])    //如果田忌最慢的比齐威王最慢的慢，田忌慢VS齐王快
                       {
                        k--;
                        sum--;
                        j++;
                    }
                    else            //田忌最慢的与齐威王最慢的相等，田忌慢VS齐王快
                       {
                        k--;
                        j++;
                    }
                }
            }
        }
        printf("%d\n",sum*200);
    }
    return 0;
}