Moving Tables-贪心

最新推荐文章于 2020-11-21 14:32:26 发布

weixin_34124577

最新推荐文章于 2020-11-21 14:32:26 发布

阅读量66

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/clnchanpin/p/6770204.html

本文探讨了一种特定场景下的桌子移动问题，并提出了两种解决该问题的算法：一种基于贪心策略，另一种则是区间覆盖的方法。通过对每种方法的具体实现和步骤的详细解释，为读者提供了理解并解决此类问题的有效途径。

Moving Tables

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Submit Status

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. 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 
     

Sample Output

      10 20 30 
     

第一种方法，很明了

对这道题目，核心算法就是贪心了。从最小的房号開始搬运。然后将不冲突的都搬走就能够了
接着就是相邻的两个房间进行特殊处理。假设是5号房间,那么6号房间的走廊就不能用。为什么。请
看题目提供的图。所以接下来就处理过去就能够了，知道没有房间能够搬。

/*
Author: 2486
Memory: 1416 KB		Time: 0 MS
Language: G++		Result: Accepted
VJ RunId: 4178448		Real RunId: 14210276
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=200+5;
int t,n;
struct obje {
    int s,t;
    bool operator<(const obje&a)const {
        return s<a.s;
    }
} objes[maxn];
bool vis[maxn];
int main() {
   // freopen("D://imput.txt","r",stdin);
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);
        for(int i=0; i<n; i++) {
            scanf("%d%d",&objes[i].s,&objes[i].t);
            if(objes[i].s>objes[i].t)swap(objes[i].s,objes[i].t);
            vis[i]=false;
        }
        sort(objes,objes+n);//按房号開始排序
        int cnt=0;
        bool flag=false;
        while(true) {
            int t=0;
            flag=false;
            for(int i=0; i<n; i++) {
                if(vis[i])continue;
                if((t%2==1&&t+1<objes[i].s)||(t%2==0&&t<objes[i].s)) {//相邻的房间处理，以及选取末尾比t大的房号
                    flag=true;
                    vis[i]=true;
                    t=objes[i].t;
                }
            }
            if(!flag)break;
            else cnt++;
        }
        printf("%d\n",cnt*10);
    }
    return 0;
}

第二种方法就是区间覆盖的方法。属于区间覆盖类型题，对于每一回合操作中记录当前覆盖的最大的值，证明这些覆盖的部分肯定不可以同一时候进行搬运，如此这里还须要一个技巧，就是将5,6或者3,4变为一个数，由于他们拥有相同的走廊。将6当做5处理。和把5当做6处理。他们都是占用同一走廊，所得到的效果是一样的

于是

/*
Author: 2486
Memory: 1408 KB		Time: 0 MS
Language: G++		Result: Accepted
 */
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=200+5;
int t,n,s,tt;
int a[maxn];
int main() {
   //freopen("D://imput.txt","r",stdin);
    scanf("%d",&t);
    while(t--) {
        memset(a,0,sizeof(a));
        scanf("%d",&n);
        int cnt=0;
        while(n--){
            scanf("%d%d",&s,&tt);
            if(s>tt)swap(s,tt);
            s=(s+1)>>1;//将他们5,6缩到一起。区间降低一半
            tt=(tt+1)>>1;
            for(int i=s;i<=tt;i++){
                a[i]++;
                if(a[i]>cnt)cnt=a[i];//求每一回合,区间的最大覆盖值就可以
            }
        }
        printf("%d\n",cnt*10);
    }
    return 0;
}

转载于:https://www.cnblogs.com/clnchanpin/p/6770204.html