Codeforces Gym 101158 C. Distribution Center

最新推荐文章于 2019-04-06 17:34:39 发布

ffutop

最新推荐文章于 2019-04-06 17:34:39 发布

阅读量1.2k

点赞数 1

CC 4.0 BY-SA版权

文章标签： codeforces

本文链接：https://blog.youkuaiyun.com/DorMOUSENone/article/details/74885803

Codeforces 同时被 2 个专栏收录

85 篇文章

订阅专栏

Gym

41 篇文章

订阅专栏

本文介绍了一种用于工厂生产线上的机器人手臂调度算法。该算法通过线性处理的方式，确定了每个储存室可以从哪些生产线接收货物，进而实现了不同生产线产品向储存室的有效混合。文章详细解释了解题思路，并提供了一个具体的C++实现示例。

题意

The factory of the Impractically Complicated Products Corporation has many manufacturing lines and the same number of corresponding storage rooms. The same number of conveyor lanes are laid out in parallel to transfer goods from manufacturing lines directly to the corresponding storage rooms. Now, they plan to install a number of robot arms here and there between pairs of adjacent conveyor lanes so that goods in one of the lanes can be picked up and released down on the other, and also in the opposite way. This should allow mixing up goods from diﬀerent manufacturing lines to the storage rooms.
Depending on the positions of robot arms, the goods from each of the manufacturing lines can only be delivered to some of the storage rooms. Your task is to ﬁnd the number of manufacturing lines from which goods can be transferred to each of the storage rooms, given the number of conveyor lanes and positions of robot arms.

解题思路

所有传送带为横向传输。将机器人按 x 坐标从小到大排序。线性处理，记录每条传送带在 x 位置前所能拿到货物的最左传送带编号和最右传送带编号，每次更新最小、最大值即可。

代码

#include<bits/stdc++.h>
using namespace std;
const int N = 200000 + 10;
const int M = 100000 + 10;
pair<int, int> p[M], l[N];
bool cmp(pair<int, int> a, pair<int, int> b) {
    return a.first < b.first;
}
int main()
{
    int n, m;
    scanf("%d %d", &n, &m);
    for(int i=1;i<=n;i++)
        l[i].first = l[i].second = i;
    for(int i=1;i<=m;i++)
        scanf("%d %d", &p[i].first, &p[i].second);
    sort(p+1, p+m+1, cmp);
    for(int i=1, y;i<=m;i++)
    {
        y = p[i].second;
        l[y].first = l[y+1].first = min(l[y].first, l[y+1].first);
        l[y].second = l[y+1].second = max(l[y].second, l[y+1].second);
    }
    for(int i=1;i<=n;i++)
        printf("%d%c", l[i].second-l[i].first+1, i==n?'\n':' ');
}