ZOJ-1029（POJ-1083、HDU-1050） Moving Tables

Moving Tables

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

Output for the Sample Input

10 
20 
30

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Source: Asia 2001, Taejon (South Korea)

————————————————————迅速的分割线————————————————————思路：一开始做了两份代码，都是错的。后来跑去看Discuss，看到了原因。其实分析题意的时候我就注意到了，例如：从1到2和从3到4是没有交集的，但是从2到3和从4到5却有交集。但是在敲代码的过程中因为拙劣的思路就放弃了这点发现。后来根据奇偶性讨论，结果错误，就只有重新思考这一点了。方法很明显，把区间转化成过道。编号为0～199即可。剩下的问题是，怎样求最小消耗时间呢？假如两个区间没有交集，那么可以划到同一个时间段内。也就是说，一个时间段内允许所有互不产生交集的区间。如果某个区间跟这个时间段内任一个区间产生了交集，那么就必须划分到另一个区间内。否则就一定可以划分到该区间内。结论就是——重合次数最多的区间的重合次数就是必须的最小时间开销。代码如下：

/****************************************/ #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <cmath> #include <stack> #include <queue> #include <vector> #include <map> #include <string> #include <iostream> using namespace std;/****************************************///只看过道，就是一个长度200的区间，被经过最大次数的区间就是最小消耗的时间//如图：横轴表示区间，纵轴表示必须花费时间的次数，次数仅取决于最多的那个//————————————————————————// ------------//-------------------// -------------  -----------//-struct Table{int from, to;}a[210];int vis[210];bool cmp(int a, int b){return a > b;}int main(){int cas;scanf("%d", &cas);while(cas--) {int n, s, t;scanf("%d", &n);for(int i = 0; i < n; i++) {scanf("%d%d", &s, &t);int u = s + t;s = min(s, t);t = u - s;a[i].from = (s + 1)>>1;a[i].to = (t + 1)>>1;}memset(vis, 0, sizeof(vis));for(int i = 0; i < n; i++) {if(a[i].from != a[i].to) {for(int k = a[i].from; k <= a[i].to; k++)vis[k]++;}elsevis[a[i].from]++;}int maxi = -1;for(int i = 0; i < 210; i++) {maxi = max(maxi, vis[i]);}printf("%d\n", maxi*10);}return 0;}