NYOJ 220 推桌子
2014-04-14 20:23
288 查看
描述
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.
输入
The input consists of T test cases. The number of test cases ) (T is given in the first line of the input file. 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 3 + N -rd
line, the remaining test cases are listed in the same manner as above.
输出
The output should contain the minimum time in minutes to complete the moving, one per line.
样例输入
样例输出
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.
输入
The input consists of T test cases. The number of test cases ) (T is given in the first line of the input file. 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 3 + N -rd
line, the remaining test cases are listed in the same manner as above.
输出
The output should contain the minimum time in minutes to complete the moving, one per line.
样例输入
3 4 10 20 30 40 50 60 70 80 2 1 3 2 200 3 10 100 20 80 30 50
样例输出
10 20 30
s ->t 但s并不一定比t小。
注意房间号到走廊号的转换 :房间号/2+房间号%2;方法不一。
计算出所有走廊所用到的的次数。找出最大的即可。
#include <stdio.h> #include <string.h> typedef struct { int s; //the table is move s to t; int t; } room; int main() { void thtime(room rm[200],int n); int i; int t; int n; int tem; room rm[200]; scanf("%d",&t); while(t--) //the cases { scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d%d",&rm[i].s,&rm[i].t); //从s到t,不一定s<t. if(rm[i].s>rm[i].t) { tem=rm[i].t; rm[i].t=rm[i].s; rm[i].s=tem; } } thtime(rm,n); } return 0; } void thtime(room rm[200],int n) { int i,j; int move[201]; int rel; memset(move,0,sizeof(move)); for (i=0;i<n;i++) { rm[i].s=rm[i].s/2+rm[i].s%2; rm[i].t=rm[i].t/2+rm[i].t%2; } for (i=0;i<n;i++) { for(j=rm[i].s;j<=rm[i].t;j++) move[j]++; } rel=move[0]; for (i=0;i<201;i++) if(move[i]>rel) rel=move[i]; printf("%d\n",rel*10); }
相关文章推荐
- 简单的四则运算
- 数的奇偶性
- ACM网址
- 1272 小希的迷宫
- 1272 小希的迷宫
- hdu 1250 大数相加并用数组储存
- 求两个数的最大公约数【ACM基础题】
- 打印出二进制中所有1的位置
- 杭电题目---一只小蜜蜂
- ACM posters
- zoj3549 快速幂
- 浅谈manacher算法 最长回文子串(Longest Palindromic Substring)
- 前缀表达式,中缀表达式,后缀表达式转化和计算
- c/c++中让输入以回车换行键结束输入
- 图论学习笔记之一——Floyd算法
- 用单调栈解决最大连续矩形面积问题
- NWERC2010 NKOJ2178 Stock Prices
- 2011ACM福州网络预选赛B题 HDU4062 Abalone
- Codeforces Round #197 (Div. 2)
- Codeforces Round #198 (Div. 1)