标签： 贪心算法

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 田忌最快的马比齐王最快的快–直接跑赢;

放上代码:

#include
#include
#include
int cmp(const void*a ,const void*b)
{
	return *(int *)b-*(int *)a;
}
int main()
{
	int i,n;
	int tian,king;
	int fa,fb,la,lb;
	while(~scanf("%d",&n)&&n!=0)
	{
		int a[1001]={0},b[1001]={0};
		getchar();
		tian=king=0;
		for(i=0;ib[fb])
			{
				tian++;
				fa++;
				fb++;
			}
			else if(a[la]>b[lb])
			{
				tian++;
				la--;
				lb--;
			}
			else if(a[la]b[fb])
			tian++;
		else if(a[fa]
	

Problem Description
FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

Input
The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1’s. All integers are not greater than 1000.

Output
For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

Sample Input
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

Sample Output
13.333
31.500

简单贪心问题。先利用结构体对J和F进行排序。然后依次累加即可。

#include 
#include 
#include 
using namespace std;

struct Room
{
	int j, f;
};

Room room[1005];

bool cmp(const Room a, const Room b)
{
	return ((double)a.j / a.f > (double)b.j / b.f);
}

int main()
{
	int m, n;
	while (cin >> m >> n && m!=-1 && n!=-1)
	{
		int i;
		for (i = 0; i > room[i].j >> room[i].f;
		}
		sort(room , room + n, cmp);
		int jsum=0, fsum = 0;
		for (i = 0; i = m) { break; }
		}
		cout  m)
		{
			fsum -= room[i].f;
			jsum -= room[i].j;
			double javasum;
			javasum = jsum+(double)room[i].j *(double(m - fsum) / room[i].f);
			cout 
	

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27554 Accepted Submission(s): 14558
Problem Description
“今年暑假不AC？”
“是的。”
“那你干什么呢？”
“看世界杯呀，笨蛋！”
“@#$%^&*%…”

确实如此，世界杯来了，球迷的节日也来了，估计很多ACMer也会抛开电脑，奔向电视了。
作为球迷，一定想看尽量多的完整的比赛，当然，作为新时代的好青年，你一定还会看一些其它的节目，比如新闻联播（永远不要忘记关心国家大事）、非常6+7、超级女生，以及王小丫的《开心辞典》等等，假设你已经知道了所有你喜欢看的电视节目的转播时间表，你会合理安排吗？（目标是能看尽量多的完整节目）

Input
输入数据包含多个测试实例，每个测试实例的第一行只有一个整数n(n<=100)，表示你喜欢看的节目的总数，然后是n行数据，每行包括两个数据Ti_s,Ti_e (1<=i<=n)，分别表示第i个节目的开始和结束时间，为了简化问题，每个时间都用一个正整数表示。n=0表示输入结束，不做处理。

Output
对于每个测试实例，输出能完整看到的电视节目的个数，每个测试实例的输出占一行。

Sample Input
12
1 3
3 4
0 7
3 8
15 19
15 20
10 15
8 18
6 12
5 10
4 14
2 9
0

Sample Output
5

动态规划+贪心，一开始想的太简单了。根本没有考虑到贪心。
放上一开始的Wa代码：

#include
int main()
{
	int t[100][2],num,n,ni,nii,time,s;
	scanf("%d",&n);
	for(ni=0;ni
放上一段比较简单的代码吧，根据节目时间长短来排序。然后依次观看时间最短的节目。(WA代码)
#include
int main()
{
	int st[100], end[100];
	int n, a, b;
	while (scanf("%d", &n) && n != 0)
	{
		for (int i = 0; iend[j])按后面排好  交换
			if (end[i]>end[j])
			{
				a = end[i];
				end[i] = end[j];
				end[j] = a;
				b = st[i];
				st[i] = st[j];
				st[j] = b;
				// 再接着开始排列
			}
		}
		int sum = 0, max = 0;
		for (int i = 0; i= max)
		{
			sum++;
			max = end[i];
		}
		printf("%dn", sum);
	}
	return 0;
}


放上已经AC的C++代码。不是我写的。
#include
#include
#include
#define MAX 1000
using namespace std;
pair p[MAX];
int main()
{
	int n;
	int s, e;
	while (scanf("%d", &n)>0 && n)
	{
		for (int i = 0; i