1018 Public Bike Management

比较复杂的模拟题，Dijstra和dfs结合，注意记牢回溯算法框架：

#include<bits/stdc++.h>
using namespace std;
#define ipair pair<int,int>
int cmax,n,sp,m;
vector<vector<int>> pre;
vector<vector<ipair>> graph;
vector<int> num,dis,temp,res;
int minNeed=INT_MAX,minBack=INT_MAX;
// 每条最短路径模拟前向再反向
void dfs(int last){
    temp.push_back(last);
    if(last==0){
        int need=0,back=0;
        for(int i=temp.size()-1;i>=0;i--){
            int tid=temp[i];
            if(num[tid]>=0){
                back+=num[tid];
            }else{ //够不够补再分情况讨论
                if(back>=(-num[tid])){ //够补
                    back+=num[tid];
                }else{    //不够补，能分多少分多少
                    need+=(-num[tid]-back);
                    back=0;
                }
            }
        }
        // 根据题目这句话：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.
        if(minNeed>need){
            minNeed=need;
            minBack=back;
            res=temp;
        }else if(minNeed==need&&minBack>back){
            minBack=back;
            res=temp;
        }
    }
    for(int i=0;i<pre[last].size();i++){
        dfs(pre[last][i]);
    }
    temp.pop_back();
}

int main(){
    scanf("%d%d%d%d",&cmax,&n,&sp,&m);
    graph.resize(n+1);
    num.resize(n+1,0);
    pre.resize(n+1);
    dis.resize(n+1,INT_MAX);
    dis[0]=0;
    for(int i=1;i<=n;i++){
        scanf("%d",&num[i]);
        num[i]-=(cmax/2);
    }
    for(int i=0;i<m;i++){
        int s,d,w;
        scanf("%d%d%d",&s,&d,&w);
        graph[s].push_back(make_pair(d,w));
        graph[d].push_back(make_pair(s,w));
    }
    priority_queue<ipair, vector<ipair>, greater<ipair>> pq;
    pq.push(make_pair(0,0)); //weight, point
    while(!pq.empty()){
        int u=pq.top().second;
        pq.pop();
        if(u==sp) break;
        for(auto& it:graph[u]){
            int v=it.first;
            int weight=it.second;
            if(dis[u]+weight<dis[v]){
                dis[v]=dis[u]+weight;
                pre[v].clear();
                pre[v].push_back(u);
                pq.push(make_pair(dis[v],v));
            }else if(dis[u]+weight==dis[v]){
                pre[v].push_back(u);
            }
        }
    }
    dfs(sp);
    printf("%d ",minNeed);
    for(int i=res.size()-1;i>=0;i--){
        printf(i==0?"%d":"%d->",res[i]);
    }
    printf(" %d",minBack);
}