技術標籤：貪心差分c貪心演算法

Problem Description

The famous ACM (Advanced Computer Maker) Company has rented a floor of a building whose shape is in the following figure.

The floor has 200 rooms each on the north side and south side along the corridor. Recently the Company made a plan to reform its system. The reform includes moving a lot of tables between rooms. Because the corridor is narrow and all the tables are big, only one table can pass through the corridor. Some plan is needed to make the moving efficient. The manager figured out the following plan: Moving a table from a room to another room can be done within 10 minutes. When moving a table from room i to room j, the part of the corridor between the front of room i and the front of room j is used. So, during each 10 minutes, several moving between two rooms not sharing the same part of the corridor will be done simultaneously. To make it clear the manager illustrated the possible cases and impossible cases of simultaneous moving.

For each room, at most one table will be either moved in or moved out. Now, the manager seeks out a method to minimize the time to move all the tables. Your job is to write a program to solve the manager’s problem.

Input

The input consists of T test cases. The number of test cases ) (T is given in the first line of the input. Each test case begins with a line containing an integer N , 1<=N<=200 , that represents the number of tables to move. Each of the following N lines contains two positive integers s and t, representing that a table is to move from room number s to room number t (each room number appears at most once in the N lines). From the N+3-rd line, the remaining test cases are listed in the same manner as above.

Output

The output should contain the minimum time in minutes to complete the moving, one per line.

Sample Input

3
4
10 20
30 40
50 60
70 80
2
1 3
2 200
3
10 100
20 80
30 50

Sample Output

10
20
30

思路

使用貪心演算法。記錄每一個通道使用的次數，使用次數最多的通道乘每次的時間即為最短時間。
下面證明一下：
假設通道使用次數最多的為N。先搬經過需要被經過N此通道的桌子（幾個不相交的區間同時搬，可能存在也為N的此時搬不了，後面證明這種情況不可能），如果這次搬之後，所有通道的次數最大為N-1，那麼再搬第二次，最大就為N-2，搬N次，最大就為0，也就是所有都為0，所有桌子都搬完了。

問題就轉換成了，證明上述第k次搬時，搬之前，所有通道需要通過的最大值M，搬之後變為M-1。假設第k次搬動的多個區間，其中一個為[a,b]而b+1這個位置沒有搬。經過b+1這個位置的有兩種情況，一種以b+1為起點，另一種情況就必須要經過b。以b+1位起點顯然不可能，如果有，那麼這個區間與[a,b]無相交，就肯定會搬。[a,b]搬完之後，所有對應的值小於等於M-1，如果此時b+1這個位置的值不小於等於M-1，所有經過b+1的必然會經過b，經過b+1的所有都搬完值為0，而b剛開始就比b+1小，最後變為負數，顯然不可能。而且，b搬過後，b+1處比M-1大，經過b+1的會經過b，b比M-1大，搬之前b比M大，與之前假設b處的值為M矛盾。得證。

另外，注意
輸入前面的數不一定大於後面的數
3-5和6-8有重疊 （5和6的通道相同）

程式碼

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

int pas[205];

int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		memset(pas,0,sizeof(pas));
		int n;
		cin>>n;
		for(int i=1;i<=n;i++)
		{
			int a,b;
			scanf("%d%d",&a,&b);
			if(a>b)
				swap(a,b);
			a=(a+1)/2;
			b=(b+1)/2;					//計算通道 
			for(int i=a;i<=b;i++)
				pas[i]++;
		}
		int maxx=0;
		for(int i=1;i<=201;i++)
			maxx=max(maxx,pas[i]);
		cout<<maxx*10<<endl;
	}
	return 0;
}

還可以用差分，時間複雜度更小

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

int pas[205];

int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		memset(pas,0,sizeof(pas));
		int n;
		cin>>n;
		for(int i=1;i<=n;i++)
		{
			int a,b;
			scanf("%d%d",&a,&b);
			if(a>b)
				swap(a,b);
			a=(a+1)/2;
			b=(b+1)/2;
			pas[a]++;				
			pas[b+1]--;
		}
		for(int i=1;i<=201;i++)
			pas[i]+=pas[i-1];
		int maxx=0;
		for(int i=1;i<=201;i++)
			maxx=max(maxx,pas[i]);
		cout<<maxx*10<<endl;
	}
	return 0;
}