pat-a1018. Public Bike Management (30)

刚开始一直没有做出来，题意没有理解对。。是在去目标地点的路上顺便使路上的车数量达到完美。我以为回来的路上也能弄完美。举例来说。a点3车，b10车，经过a到b去，要send2，back5而不是send0，back3。这样的话就比较容易做了。深搜，记录send和back的状态，找到完美路径。中间用例dijkstra算法，用来算最短路径进行剪枝，不用也不知道会不会超时。刚开始想用纯的dijkstra算法做，不知道目标到底是缺或者多，无法像路径一样每次选择最优值所以这个方法不行。

#include<cstdio>#include<map>#include<vector>#include<queue>using namespace std;int curnum[550];int dis[550];int visit[550];int vit[550];vector<int> path,ans;const int inf=10000000;struct node{int to,co;node(int a=0,int b=0):to(a),co(b){}friend bool operator < (const node& m,const node& n){return m.co>n.co;}};int c;map<int,vector<node> >g;void dij(int s){dis[s]=0;priority_queue<node> q;q.push({s,dis[s]});while(!q.empty()){node t=q.top();q.pop();int v=t.to;if(visit[v]) continue;visit[v]=1;int len=g[v].size();for(int i=0;i<len;++i){node e=g[v][i];if(dis[e.to]>dis[v]+e.co){dis[e.to]=dis[v]+e.co;q.push({e.to,dis[e.to]});}}}}int p;int mins=inf,minb=inf;void dfs(int s,int send,int back,int d){if(d>dis[p]) return;if(s==p){if(mins>send){mins=send;minb=back;ans=path;}else if(mins==send&&minb>back){minb=back;ans=path;}return;}int len=g[s].size();int a=send,b=back;for(int i=0;i<len;++i){if(vit[g[s][i].to]==0){int v=g[s][i].to;int x=curnum[v]-c/2;int y=0;if(x<0){int y=-1*x;if(back<y){send+=y-back;back=0;}else{back-=y;}}else{back+=x;}path.push_back(v);vit[v]=1;dfs(v,send,back,d+g[s][i].co);vit[v]=0;send=a,back=b;path.pop_back();}}}int main(){int n,m;int a,b,d;int send=0,back=0;scanf("%d%d%d%d",&c,&n,&p,&m);curnum[0]=c/2;for(int i=1;i<=n;++i){scanf("%d",curnum+i);dis[i]=inf;}for(int i=0;i<m;++i){scanf("%d%d%d",&a,&b,&d);g[a].push_back({b,d});g[b].push_back({a,d}); }vit[0]=1;dij(0);dfs(0,send,back,0);int len=ans.size();printf("%d",mins);printf(" 0");for(int i=0;i<len;++i) printf("->%d",ans[i]);printf(" %d",minb);return 0;} 

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