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来源：牛客网

PAT 1018  Public Bike Management

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.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 numbers: C~max~ (<= 100), always an even number, is the maximum capacity of each station; N (<= 500), the total number of stations; S~p~, 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 C~i~ (i=1,...N) where each C~i~ is the current number of bikes at S~i~ respectively. Then M lines follow, each contains 3 numbers: S~i~, S~j~, and T~ij~ which describe the time T~ij~ taken to move betwen stations S~i~ and S~j~. 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-&gtS~1~->...-&gtS~p~. Finally after another space, output the number of bikes that we must take back to PBMC after the condition of S~p~ 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.

示例1

输入

10 3 3 56 7 00 1 10 2 10 3 31 3 12 3 1

输出

3 0->2->3 0

输入:最大的容量  站台总数量  问题站台下标  路径数量

N个站台每个初始有多少个自行车

起点 终点 花费时间

首先是选择时间最短的路径，如果路径长相同，那么就选送出最少的那个，如果送出的相同，那么就选那个拿回来的最少的路径。有三个标准。

代码来自牛客网：https://www.nowcoder.com/questionTerminal/4b20ed271e864f06ab77a984e71c090f

#include <iostream>

#include <vector>

#include <limits.h>

 

using namespace std;

 

void dfs(int start, int index, int end);

 

int cmax, N, sp, M;

int costTimes, outBikes, inBikes;

int resultTimes = INT_MAX;

int resultOutBikes, resultInBikes;

vector<int> bikes, path, resultPath;

vector<vector<int> > times;

vector<bool> visited;

 

int main()

{

    ios::sync_with_stdio(false);

    // 输入数据

    cin >> cmax >> N >> sp >> M;

    bikes.resize(N+1, 0);

    visited.resize(N+1, false);

    times.resize(N+1, vector<int>(N+1, 0));//果然大佬都比较注重空间使用。

//我就直接使用501的数组了。

    for(int i=1; i<=N; i++) {

        cin >> bikes[i];

    }

    int m, n, dist;

    for(int i=0; i<M; i++) {

        cin >> m >> n >> dist;

        times[m][n] = dist;

        times[n][m] = dist;

    }

 

    // 深搜并输出结果

    dfs(0, 0, sp);

    cout << resultOutBikes << " 0";

    for(int i=1; i<resultPath.size(); i++) {

        cout << "->" << resultPath[i];

    }

    cout << " " << resultInBikes;

 

    return 0;

}

 

void dfs(int start, int index, int end)

{

    // 访问

    visited[index] = true;

    path.push_back(index);//放到路径里

    costTimes += times[start][index];

 

    // 处理

    if(index == end) {

        // 计算这条路上带去的车和带回的车

        inBikes = 0, outBikes = 0;

        for(int i=1; i<path.size(); i++) {

            if(bikes[path[i]] > cmax/2) {//如果>，那么就带回，

                inBikes += bikes[path[i]] -cmax/2;

            } else {

                if((cmax/2 - bikes[path[i]]) < inBikes) {//如果还够的话，那么就直接分配。

                    inBikes -= (cmax/2 - bikes[path[i]]);

                } else {//如果不够，那么就算到带出里

                    outBikes += (cmax/2 - bikes[path[i]]) - inBikes;

                    inBikes = 0;

                }

            }

        }

        // 判断这条路是否更好

        if(costTimes != resultTimes) {

            if(costTimes < resultTimes) {

                resultTimes = costTimes;

                resultPath = path;

                resultOutBikes = outBikes;

                resultInBikes = inBikes;

            }

        } else if(outBikes != resultOutBikes) {

            if(outBikes < resultOutBikes) {

                resultPath = path;

                resultOutBikes = outBikes;

                resultInBikes = inBikes;

            }

        } else if(inBikes < resultInBikes) {

            resultPath = path;

            resultOutBikes = outBikes;

            resultInBikes = inBikes;

        }

    } else {

        // 递归

        for(int i=1; i<=N; i++) {

            if(times[index][i] != 0 && !visited[i]) {

                dfs(index, i, end);/

            }

        }

    }

 

    // 回溯

    visited[index] = false;

    path.pop_back();

    costTimes -= times[start][index];

}

 

//反正这道题目我是不会做的。。

太厉害了。学习了。

 使用深搜，参数是当前的点，和将要访问的点，以及end点，每次进入先标记，再判断是否是end,如果不是，那么就再从当前的去循环判断；还有回溯的过程，点标记为未访问过，然后弹出路径，时间减去当前的时间。

 本来我想用的是迪杰斯特拉，但是它只能一次遍历，找不到所有的最短路径?是这样嘛?