PAT 甲级 1018. Public Bike Management (30)

题目：点击打开链接

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.
思路：
1.与1003类似，实际上是求一个最短路径的问题，我仍采用深度优先搜索；

2.路径选择的优先级是路径最短>从PBMC带走的自行车最少>带回PBMC的自行车数量最少（注意：这里的前两个条件在题目描述里出现，而最后一个则出现在输出描述，如果不注意最后一个条件将无法通过后几个测试点）；

3.注意前面站点带出来的自行车数是否满足当前站点的需要，这一点需分情况讨论。

代码：

#include<iostream>
#include<vector>
#include<climits>
using namespace std;

//需送出和需带走的自行车数量不一致
//在设计程序时,需考虑到每一个站点需要多少辆车,或是带走多少辆车

int minlen=INT_MAX;
int N;                //节点数
bool V[501]={0};      //判断该节点是否被选用
int S[501]={0};       //站点所有车辆数
int map[501][501];    //各节点之间距离
int Cmax;             //站点最大容量

int send=INT_MAX;     //需从总站携带的自行车数量
int take=0;           //从站点拿走的自行车数量

vector<int> best_r;   //选择路径

void dfs(int node,const int end,int len,vector<int> r,i
4000
nt sendnum,int takenum)
{
if(node==end)           //当前点在节点时
{
if(len<minlen || (len==minlen && sendnum<send) || (len==minlen && sendnum==send && takenum<take))      //出现新的路径
{
minlen=len;
best_r=r;       //向量复制
send=sendnum;
take=takenum;
}

return;
}
if(len>minlen)
{
return;
}

int q;
for(q=1;q<=N;++q)
{
if(V[q] && (map[node][q]!=0))
{
V[q]=false;
r.push_back(q);
if((Cmax/2-S[q]-takenum)>0)              //从之前车站带下来的自行车不足以满足要求,即根据需要额外满足
{
dfs(q,end,len+map[node][q],r,sendnum+(Cmax/2-S[q]-takenum),0);
}
else
{
dfs(q,end,len+map[node][q],r,sendnum,takenum-(Cmax/2-S[q]));
}

r.pop_back();
V[q]=true;

}
}
};

int main()
{
//输入
int i,j,k,M,p,L;
cin>>Cmax>>N>>p>>M;
for(i=1;i<=N;++i)
{
cin>>S[i];
V[i]=true;
}
for(i=0;i<M;++i)
{
cin>>j>>k>>L;
map[j][k]=L;
map[k][j]=L;
}
vector<int> R;
R.push_back(0);

dfs(0,p,0,R,0,0);
cout<<send<<" ";
for(i=0;i<best_r.size();++i)
{
cout<<best_r[i];
if(i!=(best_r.size()-1))
cout<<"->";
else
cout<<" ";
}
cout<<take<<endl;

system("pause");
return 0;
}