hdu1050Moving Tables(贪心)
Moving Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42736 Accepted Submission(s): 13986

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.
题意:有400个房间两边各200个,中间有一条走廊,在房间之间搬桌子,走廊的宽度只能容纳一张桌子,所以如果1号房间搬桌子到5号房间的时候二号房间就不能搬桌子,因为这块区域的走廊塞满了,但7号房间的走廊还是空的,可以搬,搬一次需要10分钟。问搬完给出的桌子需要多少时间。
题解:之间搬桌子的那块区域作为下标的数组值+10,每次都这样。最后这个数组中的最大值就是搬完所有桌子的时间。每次搬桌子的路径都可以看成是从低编号房间搬到高编号房间。这样方便为那块区域的数值+10,因为房间是面对面的,所以计算区域的时候可以从奇数编号房间到偶数编号房间。这样区域比较全。
#include<bits/stdc++.h>
using namespace std;
int a[];
int main() {
int t;
while(~scanf("%d",&t)) {
while(t--) {
int n;
memset(a,,sizeof(a));
scanf("%d",&n);
int maxx=;
for(int i=; i<n; i++) {
int start,end;
scanf("%d %d",&start,&end);
if(start>end)swap(start,end);
if(start%==) {
start-=;
}
if(end%)end+=;
for(int j=start; j<=end; j++) {
a[j]+=;
maxx=max(maxx,a[j]);
}
}
printf("%d\n",maxx);
} }
return ;
}
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