PAT 1018 Public Bike Management

1018 Public Bike Management

分数 30

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作者 CHEN, Yue

单位 浙江大学

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:

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: 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:

10 3 3 5
6 7 0
0 1 1
0 2 1
0 3 3
1 3 1
2 3 1

Sample Output:

3 0->2->3 0

总结：题目的难度还写不出来，继续加油！

#include 
#include 
#include 
using namespace std;
const int inf = 99999999;
int cmax, n, sp, m;
int minNeed = inf, minBack = inf;
int e[510][510], dis[510], weight[510];
bool visit[510];
vector<int> pre[510], path, temppath;
//dfs的大致流程：
//首先从目标点开始往下走，一直走到点 0，中间的路径用temppath数组存储，
//通过判断back、need、weight之间的关系来确定当前的路径否是最小需求、最小带回的路径 
void dfs(int v) {
    temppath.push_back(v);
    if(v == 0) {
        int need = 0, back = 0;
        for(int i = temppath.size() - 1; i >= 0; i--) {//从第0个点开始往回走 
            int id = temppath[i];
            if(weight[id] > 0) {//当前车辆大于完美情况，说明有车多需要带回 
                back += weight[id];
            } else {
                if(back > (0 - weight[id])) {//0-weight[id]，因为weight[id]已经确保是负数了，
				//表示当前sation需要车辆，0-weight[id]把负数转化成正数，和back比较，如果back大
				//表示从第0个点到当前点，需要带回来的车的数量大于需要带过去的车的数量 
                    back += weight[id];
                } else {
                    need += ((0 - weight[id]) - back);
                    back = 0;
                }
            }
        }
        //路径更新 
        if(need < minNeed) {
            minNeed = need;
            minBack = back;
            path = temppath;
        } else if(need == minNeed && back < minBack) {
            minBack = back;
            path = temppath;
        }
        //将最后个一点pop出来 
        temppath.pop_back();
        return ;//返回第0个点的上一个点 
    }
    for(int i = 0; i < pre[v].size(); i++)
        dfs(pre[v][i]);//查看当前点是否还有到达下一个点的路径，如果没有的话，就结束循环，最后pop出这个点 
    temppath.pop_back();
}
int main() {
    fill(e[0], e[0] + 510 * 510, inf);
    fill(dis, dis + 510, inf);
    scanf("%d%d%d%d", &cmax, &n, &sp, &m);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &weight[i]);
        weight[i] = weight[i] - cmax / 2;//整数表示有多的车，负数表示需要车 
    }
    for(int i = 0; i < m; i++) {
        int a, b;
        scanf("%d%d", &a, &b);
        scanf("%d", &e[a][b]);
        e[b][a] = e[a][b];
    }
    dis[0] = 0;
    for(int i = 0; i <= n; i++) {
        int u = -1, minn = inf;
        for(int j = 0; j <= n; j++) {
            if(visit[j] == false && dis[j] < minn) {
                u = j;
                minn = dis[j];
            }
        }
        if(u == -1) break;
        visit[u] = true;
        for(int v = 0; v <= n; v++) {
            if(visit[v] == false && e[u][v] != inf) {
                if(dis[v] > dis[u] + e[u][v]) {
                    dis[v] = dis[u] + e[u][v];
                    pre[v].clear();//将之前更新的但不是最短的路径去掉，只存最短路径 
                    pre[v].push_back(u);
                }else if(dis[v] == dis[u] + e[u][v]) {
                    pre[v].push_back(u);//路径距离相同都存下来 
                }
            }
        }
    }
    dfs(sp);
    printf("%d 0", minNeed);
    for(int i = path.size() - 2; i >= 0; i--)
        printf("->%d", path[i]);
    printf(" %d", minBack);
    return 0;
}

 好好学习，天天向上！

我要考研！