PAT1018 Public Bike Management

Public Bike Management

题目描述

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.

Figure 1

Figure 1 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:
1. 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.
2. 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.

输入描述:

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.

输出描述:

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.

输入例子:
10 3 3 5
6 7 0
0 1 1
0 2 1
0 3 3
1 3 1
2 3 1

输出例子:
3 0->2->3 0

大体思路:

找出从PBMC到Sp的最短路径，如果最短路径有多条，找从PBMC送出的车最少的，需要返回的车最少的
本题使用深度优先搜索策略较dijkstra算法要简单。
题目的解必须满足以下三个条件中的任何一个即可：
1.在所有路径中选择最短的；
2.如果路径相等，则选择从PBMC中送出最少的；
3.如果路径相等且PBMC送出的车也相等，则选择带回最少的。
另外,在实现的具体细节的时候,要注意curSend(从PBMC送出的车)与curBack(带回的车,即多余的车)的更新。
在选择点加入路径集合过程中,要把该点调整为最佳状态。
(1)如果curBack(多余的车)+该点本来就有的车数<最佳状态车数,说明需要用携带的车数来弥补，使该点达到最佳状态,同时curBack(剩余的车数)也该更新为0。
携带的车数不断累加起来就是需要从PBMC送出的车，即为curSend。
(2)如果curBack(多余的车)+该点本来就有的车数>最佳状态车数,curBack(多余的车数)就要更新=curBack(多余的车数)+该点本来就有的车数-最佳状态车数。

#include<iostream>
#include<cstring>
#include<vector>
#define INF 0x6FFFFFFF
using namespace std;
int CMax, N, Sp, M;//每个站的自行车最大容量，N个自行车，Sp终点，M条路
int map[505][505];
int vis[505];
int capacity[505];
vector<int> curPath;
vector<int> minPath;
int curCost;
int minCost;
int curSend;
int minSend;
int curBack;
int minBack;
bool isChose;
void DFS(int start)
{
    if (curCost > minCost)
    {
        return;
    }
    if (start == Sp)
    {
        isChose = false;
        if (minCost > curCost)
        {
            isChose = true;
        }
        else if (minCost == curCost)
        {
            if (minSend > curSend)
            {
                isChose = true;
            }
            else if (minSend == curSend)
            {
                if (minBack > curBack)
                {
                    isChose = true;
                }
            }
        }
        if (isChose)
        {
            minCost = curCost;
            minSend = curSend;
            minBack = curBack;
            minPath = curPath;
        }
    }
    for (int i = 1; i <= N; i++)
    {
        if (!vis[i] && map[start][i])
        {
            int lastCurCost = curCost;
            int lastCurSend = curSend;
            int lastCurBack = curBack;
            curCost += map[start][i];
            if (capacity[i]+curBack<(CMax / 2))
            {
                curSend += (CMax / 2) - (capacity[i]+curBack);
                curBack = 0;
            }
            else
            {
                curBack= (capacity[i]+curBack)-CMax/2;
            }
            vis[i] = 1;
            curPath.push_back(i);
            DFS(i);
            vis[i] = 0;
            curCost = lastCurCost;
            curSend = lastCurSend;
            curBack = lastCurBack;
            curPath.pop_back();
        }
    }
}
int main()
{
    while (scanf("%d%d%d%d",&CMax,&N,&Sp,&M)!=EOF && (CMax|| N||Sp||M))
    {
        memset(map,0,sizeof(map));
        memset(vis,0,sizeof(vis));
        memset(capacity,0,sizeof(capacity));
        curPath.clear();
        minPath.clear();
        minCost = INF;
        minSend = INF;
        minBack = INF;
        for (int i = 1; i <= N; i++)
        {
            scanf("%d",&capacity[i]);
        }
        int x, y, Cost;
        while (M--)
        {
            scanf("%d%d%d",&x,&y,&Cost);
            map[x][y] = map[y][x] = Cost;
        }
        curPath.push_back(0);
        vis[0] = 1;
        DFS(0);
        printf("%d %d",minSend,minPath[0]);
        for (int i = 1; i < minPath.size(); i++)
        {
            printf("->%d",minPath[i]);
        }
        printf(" %d\n",minBack);

    }
    system("pause");
    return 0;
}