PAT甲级1018 Public Bike Management (30)

最新推荐文章于 2022-08-16 20:45:36 发布

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最新推荐文章于 2022-08-16 20:45:36 发布

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本文链接：https://blog.youkuaiyun.com/qq_37370420/article/details/84073461

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本文介绍了一个在杭州实施的公共自行车服务，旨在为全球游客提供便利。文章详细解释了公共自行车管理中心如何通过调整各站点的自行车数量，确保它们处于理想的状态，特别是在站点出现满载或空置情况时。此外，还讨论了当问题站点被报告时，中心如何选择最优路径来解决问题，包括考虑时间和所需运送的自行车数量。

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1018 Public Bike Management （30 分）

There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any station and return it to any other stations in the city.

The Public Bike Management Center (PBMC) keeps monitoring the real-time capacity of all the stations. A station is said to be in perfect condition if it is exactly half-full. If a station is full or empty, PBMC will collect or send bikes to adjust the condition of that station to perfect. And more, all the stations on the way will be adjusted as well.

When a problem station is reported, PBMC will always choose the shortest path to reach that station. If there are more than one shortest path, the one that requires the least number of bikes sent from PBMC will be chosen.

The above figure illustrates an example. The stations are represented by vertices and the roads correspond to the edges. The number on an edge is the time taken to reach one end station from another. The number written inside a vertex S is the current number of bikes stored at S. Given that the maximum capacity of each station is 10. To solve the problem at S3, we have 2 different shortest paths:

PBMC -> S1 -> S3. In this case, 4 bikes must be sent from PBMC, because we can collect 1 bike from S1 and then take 5 bikes to S3, so that both stations will be in perfect conditions.
PBMC -> S2 -> S3. This path requires the same time as path 1, but only 3 bikes sent from PBMC and hence is the one that will be chosen.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 numbers: Cmax (≤100), always an even number, is the maximum capacity of each station; N (≤500), the total number of stations; Sp, the index of the problem station (the stations are numbered from 1 to N, and PBMC is represented by the vertex 0); and M, the number of roads. The second line contains N non-negative numbers Ci (i=1,⋯,N) where each Ci is the current number of bikes at Si respectively. Then M lines follow, each contains 3 numbers: Si, Sj, and Tij which describe the time Tij taken to move betwen stations Si and Sj. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print your results in one line. First output the number of bikes that PBMC must send. Then after one space, output the path in the format: 0−>S1−>⋯−>Sp. Finally after another space, output the number of bikes that we must take back to PBMC after the condition of Sp is adjusted to perfect.

Note that if such a path is not unique, output the one that requires minimum number of bikes that we must take back to PBMC. The judge's data guarantee that such a path is unique.

Sample Input:

Sample Output:

3 0->2->3 0

每个Si都是一个自行车车站，给定一个自行车车站最大容量cmax，那么如果自行车车站目前的容量等于cmax/2那么就是处于完美状态，如果该车站是满的或者是空的，PBMC就会携带或者从路上搜集车辆使得沿途上的自行车车站都达到完美状态，求出最短路径，如果最短路径有多个，求能带的最少的自行车数目的那条。如果还是有很多条不同的路，那么就找一个从车站带回的自行车数目最少的。

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

const int maxn=510,inf=10000000000;
int g[maxn][maxn],arr[maxn],dis[maxn];
bool vis[maxn];
vector<int> pre[maxn];
vector<int> path,temp;//temp用于临时保存路径，path用于更新最终结果
int n,c,sp,tempsend=inf,tempback=inf;

void dijkstra(int s){
	fill(dis,dis+maxn,inf);
	fill(vis,vis+maxn,false);
	dis[s]=0;
	for(int i=0;i<n;i++){
		int u=-1,minn=inf;
		for(int j=0;j<=n;j++){
			if(vis[j]==false&&dis[j]<minn){
				u=j;
				minn=dis[j];
			}
		}
		if(u==-1) return;
		vis[u]=true;
		for(int v=0;v<=n;v++){
			if(vis[v]==false&&g[u][v]!=0){
				if(dis[u]+g[u][v]<dis[v]){
					dis[v]=dis[u]+g[u][v];
					pre[v].clear();
					pre[v].push_back(u);
				}else if(dis[u]+g[u][v]==dis[v]){
					pre[v].push_back(u);
				}
			}
		}
	}
}

void dfs(int v){
	if(v==0){
		temp.push_back(v);
		int send=0,back=0;//携带数和带回数
		for(int i=temp.size()-2;i>=0;i--){
			int id=temp[i];
			if(arr[id]>c/2){
				back+=(arr[id]-c/2);//如果该站大于c/2则带回数增加
			}else{
				if(back<(c/2-arr[id])){
					//如果该站小于c/2并且带回数还不足以使该站达到c/2，携带数增加，带回数归零
					send+=(c/2-arr[id]-back);
					back=0;
				}else if(back>(c/2-arr[id])){
					back-=(c/2-arr[id]);//如果带回数足够，则减去相应数值
				}
			}
		}
		if(send < tempsend) {
			tempsend = send;
			tempback = back;
			path = temp;
		} else if(send == tempsend && back < tempback) {
			tempback = back;
			path = temp;
		}
		temp.pop_back();
		return;
	}
	temp.push_back(v);
	for(int i=0;i<pre[v].size();i++){
		dfs(pre[v][i]);
	}
	temp.pop_back();
}

int main(){
	int m,a,b,v;
	scanf("%d%d%d%d",&c,&n,&sp,&m);
	for(int i=1;i<=n;i++){
		scanf("%d",&arr[i]);
	}
	for(int i=0;i<m;i++){
		scanf("%d%d%d",&a,&b,&v);
		g[a][b]=g[b][a]=v;
	}
	dijkstra(0);
	dfs(sp);
	printf("%d 0",tempsend);
	for(int i=path.size()-2;i>=0;i--){
		printf("->%d",path[i]);
	}
	printf(" %d\n",tempback);

	return 0;
}