POJ 1083 Moving Tables(贪心)
2017-04-19 15:42
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Moving Tables
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 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.
Output
The output should contain the minimum time in minutes to complete the moving, one per line.
Sample Input
Sample Output
Source
Taejon 2001
http://poj.org/problem?id=1083
搬桌子 中间只有一条走廊 每次搬都要进过中间 有时就必须出现等待的情况
1 3 2 200就必须等一个搬完再搬第二次了 最好的大家一起搬没矛盾的情况下
最少多少时间搬完 每一次10分钟
我的想法是贪心
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int vis[1010];
struct node
{
int x,y;
} aa[1010];
bool cmp(node a,node b)
{
if(a.x==b.x)
return a.y<b.y;
return a.x<b.x;
}
int main()
{ int t,u,v,k,i,j,sum,p;
scanf("%d",&t);
while(t--)
{
scanf("%d",&k);
for(i=0; i<k; i++)
{
scanf("%d %d",&u,&v);
if(v<u) swap(u,v);
if(u%2==0) u--;
if(v%2==0) v--;//偶全变成相应的奇数
aa[i].x=u,aa[i].y=v;
}
sort(aa,aa+k,cmp);
memset(vis,0,sizeof(vis));
sum=0;
for(i=0; i<k; i++)//看最多一个能搬多少桌子
{
if(vis[i]) continue;
p=aa[i].y;
for(j=i+1; j<k; j++)
{
if(vis[j]) continue;
if(aa[j].x>p)
{
vis[j]=1;
p=aa[j].y;
}
}
sum++;
}
printf("%d\n",sum*10);
}
return 0;
}
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 31715 | Accepted: 10591 |
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 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.
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
Source
Taejon 2001
http://poj.org/problem?id=1083
搬桌子 中间只有一条走廊 每次搬都要进过中间 有时就必须出现等待的情况
1 3 2 200就必须等一个搬完再搬第二次了 最好的大家一起搬没矛盾的情况下
最少多少时间搬完 每一次10分钟
我的想法是贪心
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int vis[1010];
struct node
{
int x,y;
} aa[1010];
bool cmp(node a,node b)
{
if(a.x==b.x)
return a.y<b.y;
return a.x<b.x;
}
int main()
{ int t,u,v,k,i,j,sum,p;
scanf("%d",&t);
while(t--)
{
scanf("%d",&k);
for(i=0; i<k; i++)
{
scanf("%d %d",&u,&v);
if(v<u) swap(u,v);
if(u%2==0) u--;
if(v%2==0) v--;//偶全变成相应的奇数
aa[i].x=u,aa[i].y=v;
}
sort(aa,aa+k,cmp);
memset(vis,0,sizeof(vis));
sum=0;
for(i=0; i<k; i++)//看最多一个能搬多少桌子
{
if(vis[i]) continue;
p=aa[i].y;
for(j=i+1; j<k; j++)
{
if(vis[j]) continue;
if(aa[j].x>p)
{
vis[j]=1;
p=aa[j].y;
}
}
sum++;
}
printf("%d\n",sum*10);
}
return 0;
}
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