SDAU课程练习--problemC

题目描述

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.

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

Sample Output

题目大意

田忌与king赛马，在赛马过程中使用贪心。

思路

将总的问题分解为一个个小问题，每个小问题都选择最佳，最佳的选择一定是从田忌的最快最慢与king的最快最慢中选择对决马匹。分别记为tf，ts，kf，ks。分三种情况：

tf>kf 此时选择tf与kf打是最优的。因为tf反正都要赢，最好去赢对方最快的
tf<kf 此时选择ts与kf的是最优的。因为kf反正都要赢，就让我方最慢的消耗掉它
tf=kf
1. ts>ks 此时选择ts与ks打。因为ks反正要输，让我方最慢的和它打为最优选择
2. ts<ks 此时选择ts与kf打。因为ts反正都要输，不如输给对方最快的
3. ts=ks
  1. ts<kf 用ts与kf打，消耗对方最快的。
  2. ts=kf 游戏结束 game over
  3. ~~ts>kf~~ 不可能情况

在做题过程中出现了一个插曲，我写的代码怎么也过不了。我在一个博客中找到的代码和我思路差不多（比较最慢的）

用他的代码一边就A了。真是见了鬼了

下面贴出两个代码

AC代码

    #include<stdio.h>
    #include<algorithm>
    using namespace std;
    int main(){
        int n, i, j;
        int a[1000];
        int b[1000];
        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);
            sort(b, b + n);
            int c = 0, d = 0, a = n - 1, b = n - 1, ans = 0;
            for(i = 0; i < n; ++i){
                if(a[c] > b[d]){
                    ++c;
                    ++d;
                    ++ans;
                }else if(a[c] < b[d]){
                    ++c;
                    --b;
                    --ans;
                }else if(a[a] > b[b]){
                    --a;
                    --b;
                    ++ans;
                }else if(a[a] < b[b]){
                    ++c;
                    --b;
                    --ans;
                }else if(a[c] < b[b]){
                    ++c;
                    --b;
                    --ans;
                }else{
                    break;
                }
            }
            printf("%d\n", ans * 200);
        }
        return 0;
    }

我的代码

#include<iostream>
#include<algorithm>
using namespace std;
bool cmp(int &e, int &f)
{
    return e > f;
}
int main()
{
    //freopen("date.in", "r", stdin);
    //freopen("date.out", "w", stdout);
    int *a ,*b,*c, *d;//a c为田最大最小，b，d为王最大最小
    int t[1000], w[1000];
    int jie=0,N;
    while (cin>>N&&N)
    {
        jie = 0;
        for (int i = 0; i < N; i++)
            cin >> t[i];
        for (int i = 0; i < N; i++)
            cin >> w[i];
        sort(t, t + N,cmp);
        sort(w, w + N,cmp);
        a = t;
        c = t+N-1;
        b= w;
        d= w+N - 1;
        for (int i = 0; i<N;i++)
        {
            if (*a > *b)
            {
                jie++;
                a++;
                b++;
            }
            else
            {
                if (*a < *b)
                {
                    jie--;
                    c--;
                    b++;
                }
                else
                    if (*c < *d)
                    {
                        jie--;
                        c--;
                        b++;
                    }
                    else
                        if (*c > *d)
                        {
                            jie++;
                            c--;
                            d--;
                        }
                        else
                            if (*c < *b)
                            {
                                jie--;
                                c--;
                                b++;
                            }
                            else
                            {
                                jie = 0;
                                goto shan;
                            }

            }
        }
        shan:cout << jie * 200 << endl;
    }
}

真是见鬼了，这两段代码耗费了我一个星期六。。

先贴在这里，过两天再研究，先往下刷题吧！

来自为知笔记(Wiz)