题目：
Tian Ji -- The Horse Racing |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 4411 Accepted Submission(s): 1069 |
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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. |
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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. |
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Output For each input case, output a line containing a single number, which is the maximum money Tian Ji will get, in silver dollars. |
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Sample Input 3 92 83 71 95 87 74 2 20 20 20 20 2 20 19 22 18 0 |
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Sample Output 200 0 0 |
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Source 2004 Asia Regional Shanghai |
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Recommend JGShining |
题目大意：
这么长的一段扯淡的话，也是醉了。不过这也是体现实力的一部分啊。当时去成都现场赛的时候，记得有一道题扯了1页半纸，最后的一丝就是让你把空格键换成Tab键输出。。想想也是醉了。。。
题目分析：
一道简单的贪心的题目，挺经典的。田忌赛马这道题可以用很多种做法来做：贪心、DP、二分图的最大匹配什么的都可以做的。以下介绍一下用贪心来怎么做这道题。贪心策略如下：
1、如果田忌最慢的马比齐王最慢的马快，则比之。
2、如果田忌最慢的马比齐王最慢的马慢，则将田忌最慢的马和齐王最快的马比。//贪心的第一步.（横竖都是输，和最快的比能为后面创造有利的局势。如田忌：10 20 80 齐王： 11 30 80 。）
3、如果田忌最慢的马和齐王最慢的马速度一样。则进行以下分析
1）田忌最快的马比齐王最快的马要快，则比之
2）田忌最快的马比齐王最快的马要慢，则用田忌最慢的马比齐王最快的马。(为后面创造有利的局势)
3）田忌最快的马和齐王最快的马速度一样，则用田忌最慢的马比齐王最快的马。
当然这道题倒过来考虑也是可以的，也就是从快的开始分析也是可以的。
代码如下：
/*
* c.cpp
*
* Created on: 2015年1月29日
* Author: Administrator
*/
#include <iostream>
#include <cstdio>
#include <algorithm>//sort()函数在这个头文件里
using namespace std;
const int maxn = 1005;
int a[maxn];
int b[maxn];
int main(){
int n;
while(scanf("%d",&n)!=EOF,n){
int i;
for(i = 0 ; i < n ; ++i){
scanf("%d",&a[i]);
}
for(i = 0 ; i < n ; ++i){
scanf("%d",&b[i]);
}
//对数据进行排序,利用有序化数据尽心贪心选择.体现贪心思想
sort(a,a+n);
sort(b,b+n);
int tl = 0;//用来标记田忌目前最慢的马
int ql = 0;
int tr = n-1;//用来标记天际目前最快的马
int qr = n-1;
int sum = 0;//最后田忌可以争取的金钱
while(tl <= tr){//如果还没有比完
if(a[tl] > b[ql]){//如果田忌最慢的马比齐王最慢的马要快
//则田忌慢VS齐王慢
sum += 200;
tl++;
ql++;
}else if(a[tl] == b[ql]){//如果田忌最慢的马==齐王最慢的马
while(tl <= tr && ql <= qr){//在还没有结束的情况下不断地进行比较
if(a[tr] > b[qr]){//如果田忌最快的马比齐王最快的马要快
//则田忌快VS齐王快
sum += 200;
tr--;
qr--;
}else{//在田忌最快的马比齐王最快的马要慢的情况下都是田忌慢VS齐王快
if(a[tl] < b[qr]){//田忌最慢的马不一定要比齐王最快的马要慢。例如双方所有马匹的速度都是一样的
sum -= 200;
}
tl++;
qr--;
break;
}
}
}else{//如果田忌最慢的马比齐王最慢的马慢
//则田忌慢VS齐王快
sum -= 200;
tl++;
qr--;
}
}
printf("%d\n",sum);
}
return 0;
}