1018. Public Bike Management (30)

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->S₁->...->S_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.

Sample Input:

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

Sample Output:

3 0->2->3 0

首先DFS找最短路径，寻找途中记录需要从总站发送的车和当前总共多余的车即用于补充下一站或者最终送回总站的车，例如4 10需要send为1back为5；

如果直接记录总共需要的车也只有2个测试点错误，即4 10记录成send -4，感觉真的考试可能没办法想到这么多了。

一开始用数组记录每个站点的Send的车不知道为什么也有2个测试点会有问题，仍旧未解决，

#include <iostream>#include <vector>#include <iterator>#include <utility>#include <cstdio>#define MAX 505#define INF (0x7fffffff)using namespace std;struct Node{    bool known;    int path;    int bike;    vector<pair<int, int> > v;};Node graph[MAX] = {false};vector<int> save;int dist[MAX] = {0};int Cmax, N, Sp, M;int mindist = INF, minsend = INF, minback = INF;int Back = 0, Send = 0;void init(){    int bikenum;    int v1, v2, w;    cin >> Cmax >> N >> Sp >> M;    for(int i=1; i<=N; i++)    {        cin >> bikenum;        graph[i].bike = bikenum;    }    for(int i=0; i<M; i++)    {        cin >> v1 >> v2 >> w;        graph[v1].v.push_back(make_pair(v2, w));        graph[v2].v.push_back(make_pair(v1, w));    }    graph[0].known = true;    graph[0].path = -1;}void update(int x){    save.clear();    minback = Back;    mindist = dist[x];    minsend = Send;    while(x!=-1)    {        save.push_back(x);        x = graph[x].path;    }}void DFS(int x){    if(dist[x] > mindist)        return ;    if(x == Sp)    {        if(dist[x] < mindist)//距离较小则更新        {            update(x);        }        else if(dist[x] == mindist)        {            if(Send < minsend || (Send==minsend && Back<minback))//距离相等则Send较小更新，Send相等则Back较小更新                update(x);        }        return ;    }    vector<pair<int, int> >::iterator it;    for(it=graph[x].v.begin(); it!=graph[x].v.end(); it++)    {        if(!graph[it->first].known)        {            graph[it->first].known = true;            graph[it->first].path = x;//记录路径            int curBack = Back;//记录当前back和send用于回溯            int curSend = Send;            if(graph[it->first].bike < Cmax/2)            {                Back -= Cmax/2-graph[it->first].bike;//如果当前站车小于一半看之前back的车够不够，如果够就不需要从总站补充，如果不够则需要补充Send                if(Back < 0)                {                    Send -= Back;                    Back = 0;                }            }            else{                Back += graph[it->first].bike-Cmax/2;//如果当前站大于一半，则多余的用于Back            }            dist[it->first] = dist[x]+it->second;//记录路长            DFS(it->first);            graph[it->first].known = false;            Back = curBack;            Send = curSend;        }    }}int main(){    //freopen("in.txt", "r", stdin);    init();    DFS(0);    cout << minsend << " ";    for(int i=save.size()-1; i>=0; i--)    {        if(i==0)            cout << save[i] << " ";        else            cout << save[i] << "->";    }    cout << minback << endl;    return 0;}