【PAT1018】 Public Bike Management 单源最短路径&路径记录回溯
1018. Public Bike Management (30)
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: 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 1Sample Output:
3 0->2->3 0
给一张图, 每个node都代表一个杭州的一个借或者还自行车站点, node上的值表示当前这个站点拥有多少量自行车, 每条边表示两个站点之间要花多少时间从一个站点到另一个站点, 给定一个有问题的站点, 求出从控制中心(PBMC)到该站点的最短路径并且使得带出去以及拿回来的自行车的数量最少.
分析：
①用最短路径算法(比如迪杰斯特拉)得到从PBMC到问题站点的所有最短路径；
②在Dijstra算法中，记录下每个节点的最优路径的上一节点，若有多个则全部记录下来；
③根据要求（带来最少带回最少），从目标节点的多个相同最短路径中找到最满足要求的一个；
④根据路径中记录的上一节点，进行回溯，直到根节点结束。
代码：
#include <iostream>
#include <fstream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <stack>
using namespace std;
//此代码使用前，需删除下面两行+后面的system("PAUSE")
ifstream fin("in.txt");
#define cin fin
#define MAXNUM 501
int curNum[MAXNUM]={0}; //每个station的自行车数量
int link[MAXNUM][MAXNUM]={0};
int minSumPath[MAXNUM]={0}; //达到每个station的最短路径长度
bool visited[MAXNUM]={false};
const int INF = 0x7fffffff;
struct LastNode{
int lastid; //上一个节点
int bikeSum; //到当下的车辆总数
int staCount; //到当下的站点总数
LastNode(int l,int b,int s){lastid=l;bikeSum=b;staCount=s;}
};
vector<LastNode> vec[MAXNUM];
int c,n,s,m;
void push(int target,int dataid){ //把dataid节点的统计数据导入到target节点
for(int i=0;i<vec[dataid].size();i++){
vec[target].push_back(LastNode(dataid,vec[dataid][i].bikeSum+curNum[target],vec[dataid][i].staCount+1));
}
}
void Dij(int source,int target){
int cen = source;
vec[0].push_back(LastNode(-1,0,0));
int min,minIndex,newSumPath;
while(1){
visited[cen]=true;
min = INF;
for(int i=1;i<=n;i++){
if(visited[i])continue;
if(link[cen][i]!=0){
newSumPath = minSumPath[cen]+link[cen][i];
if(minSumPath[i]==0 || minSumPath[i]>newSumPath){ //有更小路径，更新
minSumPath[i] = newSumPath;
vec[i].clear(); //先清空vector再导入
push(i,cen);
}else if(minSumPath[i] == newSumPath){ //路径相等，添加
push(i,cen); //再原有数据基础上添加
}
}
if(min > minSumPath[i] && minSumPath[i]!=0){
min = minSumPath[i];
minIndex = i;
}
}
if(minIndex == target){ //如果找到目标节点，直接跳出循环
break;
}
cen = minIndex;
}
}
stack<int> handleIndex(int index){ //根据目标节点回溯到0号节点，将路径存储在stack中
stack<int> res;
int id = index;
int cen = s;
int staCount,bikeSum;
int i;
while(1){
staCount = vec[cen][id].staCount-1;
bikeSum = vec[cen][id].bikeSum - curNum[cen];
res.push(cen);
cen = vec[cen][id].lastid;
if(cen == -1)break; //回溯到根节点 则跳出循环
for(i=0;i<vec[cen].size();i++){
if(vec[cen][i].staCount == staCount && vec[cen][i].bikeSum == bikeSum){ //看上一个节点的哪条数据满足条件
id = i;
break;
}
}
}
return res;
}
int main()
{
cin>>c>>n>>s>>m;
if(n==0 || s==0){
cout<<"0 0 0"<<endl;
return 0;
}
int i;
for(i=0;i<n;i++)cin>>curNum[i+1];
int a,b;
for(i=0;i<m;i++){
cin>>a>>b;
cin>>link[a][b];
link[b][a]=link[a][b];
}
Dij(0,s);
int plusZero = INF;
int plusIndex = -1;
int minusZero = -1000000;
int minusIndex = -1;
int index = -1;
int need; //需要带来的车辆数
vector<LastNode> tt = vec[s];
for(i=0;i< tt.size();i++){ //在目标节点 所有相同的最短路径中 找带来最少 带回最少的路径下标
need = tt[i].staCount*c/2 - tt[i].bikeSum;
if(need > 0){
if(need < plusZero){
plusZero = need;
plusIndex = i;
}
}else if(need < 0){
if(need > minusZero){
minusZero = need;
minusIndex = i;
}
}else{
index = i;
}
}
stack<int> res;
int come,back;
if(index != -1){
res = handleIndex(index); //根据目标节点回溯到0号节点
come = 0;
back = 0;
}else if(plusIndex != -1){
res = handleIndex(plusIndex);
come = plusZero;
back = 0;
}else{
res = handleIndex(minusIndex);
come = 0;
back = 0-minusZero;
}
cout<<come<<' ';
cout<<res.top(); //输出路径
res.pop();
while(!res.empty())
{
cout<<"->"<<res.top();
res.pop();
}
cout<<" "<<back<<endl;
system( "PAUSE");
return 0;
}
结果：
| 测试点 | 结果 | 用时(ms) | 内存(kB) | 得分/满分 |
|---|---|---|---|---|
| 0 | 答案正确 | 2 | 380 | 12/12 |
| 1 | 答案正确 | 2 | 372 | 2/2 |
| 2 | 答案正确 | 2 | 376 | 2/2 |
| 3 | 答案正确 | 2 | 376 | 2/2 |
| 4 | 答案正确 | 1 | 380 | 2/2 |
| 5 | 答案错误 | 2 | 256 | 0/2 |
| 6 | 答案正确 | 2 | 376 | 2/2 |
| 7 | 答案错误 | 2 | 376 | 0/3 |
| 8 | 答案正确 | 2 | 376 | 2/2 |
| 9 | 答案正确 | 5 | 1404 | 1/1 |
还有两个Case不过，没发现原因，就先放这里吧。