poj1083 贪心
Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
Description
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
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.
Output
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
题目大意:有400个房屋,需要在这些房屋之间搬运箱子,但是同一段走廊在每10分钟只能进行一次搬运,问你最少要多少分钟才能完成搬运任务。
思路分析:我看题目分类这道题是划分在dp里面的,刚开始一直想构造状态转移方程,就是没有弄出来,后来看了别人的题解,才造用贪心做,直接
记录每个区间的访问次数,访问最大次数*10就是需要的最短时间。
代码:
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
const int maxn=400+10;
const int inf=0xffffff;
int t[maxn];
int main()
{
int T,n,a,b;
cin>>T;
while(T--)
{
scanf("%d",&n);
mem(t);
int ma=-inf;
for(int i=0;i<n;i++)
{
scanf("%d%d",&a,&b);
if(a>b) swap(a,b);
if(a%2==0) a--;
if(b%2==1) b++;
for(int j=a;j<=b;j++)
t[j]++;
}
for(int i=1;i<=400;i++)
{
ma=max(ma,t[i]);
}
cout<<10*ma<<endl;
}
return 0;
}
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