PAT甲級1018 Public Bike Management (30)
1018 Public Bike Management (30 分)
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
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:
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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.
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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
每個Si都是一個自行車車站,給定一個自行車車站最大容量cmax,那麼如果自行車車站目前的容量等於cmax/2那麼就是處於完美狀態,如果該車站是滿的或者是空的,PBMC就會攜帶或者從路上搜集車輛使得沿途上的自行車車站都達到完美狀態,求出最短路徑,如果最短路徑有多個,求能帶的最少的自行車數目的那條。如果還是有很多條不同的路,那麼就找一個從車站帶回的自行車數目最少的。
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn=510,inf=10000000000;
int g[maxn][maxn],arr[maxn],dis[maxn];
bool vis[maxn];
vector<int> pre[maxn];
vector<int> path,temp;
int n,c,sp,tempsend=inf,tempback=inf;
void dijkstra(int s){
fill(dis,dis+maxn,inf);
fill(vis,vis+maxn,false);
dis[s]=0;
for(int i=0;i<n;i++){
int u=-1,minn=inf;
for(int j=0;j<=n;j++){
if(vis[j]==false&&dis[j]<minn){
u=j;
minn=dis[j];
}
}
if(u==-1) return;
vis[u]=true;
for(int v=0;v<=n;v++){
if(vis[v]==false&&g[u][v]!=0){
if(dis[u]+g[u][v]<dis[v]){
dis[v]=dis[u]+g[u][v];
pre[v].clear();
pre[v].push_back(u);
}else if(dis[u]+g[u][v]==dis[v]){
pre[v].push_back(u);
}
}
}
}
}
void dfs(int v){
if(v==0){
temp.push_back(v);
int send=0,back=0;
for(int i=temp.size()-2;i>=0;i--){
int id=temp[i];
if(arr[id]>c/2){
back+=(arr[id]-c/2);
}else{
if(back<(c/2-arr[id])){
send+=(c/2-arr[id]-back);
back=0;
}else if(back>(c/2-arr[id])){
back-=(c/2-arr[id]);
}
}
}
if(send < tempsend) {
tempsend = send;
tempback = back;
path = temp;
} else if(send == tempsend && back < tempback) {
tempback = back;
path = temp;
}
temp.pop_back();
return;
}
temp.push_back(v);
for(int i=0;i<pre[v].size();i++){
dfs(pre[v][i]);
}
temp.pop_back();
}
int main(){
int m,a,b,v;
scanf("%d%d%d%d",&c,&n,&sp,&m);
for(int i=1;i<=n;i++){
scanf("%d",&arr[i]);
}
for(int i=0;i<m;i++){
scanf("%d%d%d",&a,&b,&v);
g[a][b]=g[b][a]=v;
}
dijkstra(0);
dfs(sp);
printf("%d 0",tempsend);
for(int i=path.size()-2;i>=0;i--){
printf("->%d",path[i]);
}
printf(" %d\n",tempback);
return 0;
}