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A1018 Public Bike Management (30分)

作者:互联网

一、技术总结

  1. 这一题,题意的理解十分关键,主要有两个点,在花费时间最少的前提下,一个是在去的路上进行调整,能够带的单车最少优先;如果还是有多条,那么带回单车最少的优先。说明只能在去的路上对车站的车进行调整,回的时候不能够进行调整,试想如果回的时候也可以调整,那么就不会出现第二个限制条件了。
  2. 使用Djikstra+DFS
  3. 同时在每个车站的车数量输入,进行减去Cmax/2,直接对于操作,方便后续操作。具体参考代码处
  4. 对于每个车站多了两个属性,一个是need,一个是remain,need记录到该车站处所欠缺的数量,也就是需要从管理中心调过来车的数量;remain记录到该车站处,多余的车,也就是往回运的车。

二、参考代码

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 510;
const int INF = 0x3fffffff;
int G[MAXN][MAXN];
int d[MAXN], weight[MAXN], w[MAXN];
vector<int> path, tempPath;
vector<int> pre[MAXN];
bool vis[MAXN] = {false};
int Cmax, n, s, m;
int minNeed = INF, minRemain = INF;
void Djikstra(int s){
    fill(d, d+MAXN, INF);
    d[s] = 0;
    for(int i = 0; i <= n; i++){
        int u = -1, MIN = INF;
        for(int j = 0; j <= n; j++){
            if(vis[j] == false && d[j] < MIN){
                u = j;
                MIN = d[j];
            }
        }
        if(u == -1) return;
        vis[u] = true;
        for(int v = 0; v <= n; v++){
            if(vis[v] == false && G[u][v] != INF){
                if(d[u] + G[u][v] < d[v]){
                    d[v] = d[u] + G[u][v];
                    pre[v].clear();
                    pre[v].push_back(u);
                }else if(d[u] + G[u][v] == d[v]){
                    pre[v].push_back(u);
                }
            }
        }
    }
}
void DFS(int v){
    if(v == 0){
        tempPath.push_back(v);
        int need = 0, remain = 0;
        for(int i = tempPath.size()-1; i >= 0; i--){
            int id = tempPath[i];
            if(weight[id] > 0){
                remain += weight[id];
            }else{
                if(remain > abs(weight[id])){
                    remain -= abs(weight[id]);
                }else{
                    need += abs(weight[id]) - remain;
                    remain = 0;
                }
            }

        }
        if(need < minNeed){//优化,如果所需的更少 
            minNeed = need;
            minRemain = remain;
            path = tempPath; 
        }else if(need == minNeed && remain < minRemain){
            //携带数目相同,带回数目变多
            minRemain = remain;
            path = tempPath; 
        }
        tempPath.pop_back();
        return;
    }
    tempPath.push_back(v);
    for(int i = 0; i < pre[v].size(); i++){
        DFS(pre[v][i]);
    }
    tempPath.pop_back();
}
int main(){
    scanf("%d%d%d%d", &Cmax, &n, &s, &m);
    fill(G[0], G[0]+MAXN*MAXN, INF);
    //weight[0] = 0;
    for(int i = 1; i <= n; i++){
        scanf("%d", &weight[i]);
        weight[i] -= Cmax/2;
    }
    int id1, id2, L;
    for(int i = 0; i < m; i++){
        scanf("%d%d%d", &id1, &id2, &L);
        G[id1][id2] = L;
        G[id2][id1] = L;
    }
    Djikstra(0);
    DFS(s);
    int number = path.size()-1;
    printf("%d ", minNeed);
    for(int i = path.size()-1; i >= 0; i--){
        printf("%d", path[i]);
        if(i > 0) printf("->");
    }
    printf(" %d", minRemain);
    return 0;
}

标签:Management,weight,int,remain,Bike,A1018,tempPath,need,MAXN
来源: https://www.cnblogs.com/tsruixi/p/12403147.html