题干：
链接：https://www.nowcoder.com/questionTerminal/4b20ed271e864f06ab77a984e71c090f
来源：牛客网
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.
输入描述:
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.
输出描述:
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.示例1
输入
10 3 3 5
6 7 0
0 1 1
0 2 1
0 3 3
1 3 1
2 3 1输出
3 0->2->3 0再给一组样例：
100 6 1 15
78 54 0 37 36 82
0 1 7
0 2 4
0 3 7
0 4 7
0 5 2
1 2 9
1 3 1
1 4 2
1 5 5
2 3 1
2 4 2
2 5 7
3 4 4
3 5 7
4 5 7
对应输出应该为:
46 0->2->3->1 28
题目大意：每个自行车车站的最大容量为一个偶数cmax，如果一个车站里面自行车的数量恰好为cmax / 2，那么称处于完美状态。如果一个车站容量是满的或者空的，控制中心（处于结点0处）就会携带或者从路上收集一定数量的自行车前往该车站，一路上会让所有的车站沿途都达到完美。现在给出cmax，车站的数量n，问题车站sp，m条边，还有距离，求最短路径。如果最短路径有多个，求能带的最少的自行车数目的那条。如果还是有很多条不同的路，那么就找一个从车站带回的自行车数目最少的。带回的时候是不调整的
解题报告：
这题解法多种多样，此处列举二三。
AC代码1：（双dfs）
#include<bits/stdc++.h>using namespace std;
const int INF = 0x3f3f3f3f;
int maze[505][505];
int val[505],dis[505];
bool vis[505];
int ans;
int c,n,s,m;
void Dijkstra() {memset(dis,INF,sizeof dis);memset(vis,0,sizeof vis);dis[0]=0;int all = n+1,minw,minv;while(all--) {minw = INF;for(int i = 0; i<=n; i++) {if(vis[i]) continue;if(dis[i] < minw) {minw = dis[i];minv = i;}}vis[minv] = 1;for(int i = 0; i<=n; i++) {if(vis[i] || maze[minv][i] == INF) continue;dis[i] = min(dis[i],dis[minv] + maze[minv][i]);}}
}
void dfs(int x,int res) {if(res < 0) return ;if(x == s) {ans = min(ans,res);return ;}for(int i = 0; i<=n; i++) {if(maze[x][i] == INF) continue;if(dis[i] == dis[x] + maze[x][i]) dfs(i,res - (c-val[i]));}
}
bool Dfs(int x, int y, int z) {if (x==0) {if (y == z) {printf("0");return 1;} else return 0;}for (int i = 0; i <= n; i++) {if (maze[x][i] == -1) continue;if (dis[x] == dis[i] + maze[x][i]) {if (Dfs(i, y + c - val[x], z)) {printf("->%d", x);return 1;}}}return 0;
}int main() {int a,b,w;cin>>c>>n>>s>>m;c>>=1;for(int i = 1; i<=n; i++) scanf("%d",val+i);memset(maze,INF, sizeof maze);for(int i = 1; i<=m; i++) {scanf("%d%d%d",&a,&b,&w);if(w < maze[a][b]) maze[a][b] = maze[b][a] = w;}Dijkstra();int l = 0,r = n*c;int mid = (l+r)/2;while(l < r) {mid = (l+r)/2;ans = INF;dfs(0,mid);if(ans == INF) l = mid+1;else r=mid;}printf("%d ",l);dfs(0,l);Dfs(s,ans,l);printf(" %d\n", ans);return 0 ;
}此法先用邻接矩阵存图，然后跑Dijkstra求出1号点到各点的最短路，然后二分答案从中心带出多少辆车，顺便求出需要带回多少辆车，因为数据保证有且唯一正确的路径，所以倒序找到路径，一定能找到、、、、总之不是很好理解。
AC代码2：
#include<bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
int cmax, n, sp, m;
int minNeed = inf, minBack = inf;
int e[510][510], dis[510], weight[510];
bool vis[510];
vector<int> pre[510];
vector<int> path, temppath;
void dfs(int v) {temppath.push_back(v);if(v == 0) {int need = 0, back = 0;for(int i = temppath.size() - 1; i >= 0; i--) {int id = temppath[i];if(weight[id] > 0) {back += weight[id];} else {if(back > (0 - weight[id])) {back += weight[id];} else {need += ((0 - weight[id]) - back);back = 0;}}}if(need < minNeed) {minNeed = need;minBack = back;path = temppath;} else if(need == minNeed && back < minBack) {minBack = back;path = temppath;}temppath.pop_back();return ;}for(int i = 0; i < pre[v].size(); i++)dfs(pre[v][i]);temppath.pop_back();
}
int main() {memset(e,inf,sizeof e);memset(dis,inf,sizeof dis);scanf("%d%d%d%d", &cmax, &n, &sp, &m);for(int i = 1; i <= n; i++) {scanf("%d", &weight[i]);weight[i] = weight[i] - cmax / 2;}for(int i = 0; i < m; i++) {int a, b;scanf("%d%d", &a, &b);scanf("%d", &e[a][b]);e[b][a] = e[a][b];}dis[0] = 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) break;vis[u] = true;for(int v = 0; v <= n; v++) {if(vis[v] == false && e[u][v] != inf) {if(dis[v] > dis[u] + e[u][v]) {dis[v] = dis[u] + e[u][v];pre[v].clear();pre[v].push_back(u);} else if(dis[v] == dis[u] + e[u][v]) {pre[v].push_back(u);}}}}dfs(sp);printf("%d 0", minNeed);for(int i = path.size() - 2; i >= 0; i--)printf("->%d", path[i]);printf(" %d", minBack);return 0;
}这种方法相对来说较好理解，就是先dijkstra跑出最短路条数并且记录路径，然后dfs搜索出最优路径
分析：Dijkstra + DFS。如果只有Dijkstra是不可以的，因为minNeed和minBack在路径上的传递不满足最优子结构，不是简单的相加的过程，只有在所有路径都确定了之后才能去选择最小的need和最小的back。
Dijkstra求最短路径，dfs求minNeed和minBack和path，dfs的时候模拟一遍需要调整的过程，求出最后得到的need和back，与minNeed和minBack比较然后根据情况更新path，最后输出minNeed path 和 minBack，记得path是从最后一个结点一直到第一个结点的，所以要倒着输出~
AC代码3：
这个更不知道为什么了。。甚至回溯都没看懂，为啥注释那一行加上，for中那一行去掉，就不对了
#include <bits/stdc++.h>
using namespace std;
const int maxn=500+5,INF=1e9;
int cap,n,sp,m,bike[maxn],cost[maxn][maxn];
vector<int> G[maxn],anspath,curpath;
int minsend=INF,minback=INF,mincost=INF,vis[maxn];
void dfs(int cur,int cursend,int curback,int curcost){vis[cur]=1;curpath.push_back(cur);if(curcost>mincost) return ;if(cur==sp){if(curcost<mincost){mincost=curcost;minsend=cursend;minback=curback;anspath=curpath;return ;}if(curcost==mincost&&cursend<minsend){mincost=curcost;minsend=cursend;minback=curback;anspath=curpath;return ; }if(curcost==mincost&&cursend==minsend&&curback<minback){mincost=curcost;minsend=cursend;minback=curback;anspath=curpath;return ;}return ;}for(int i=0;i<G[cur].size();++i){int v=G[cur][i];if(!vis[v]){if(bike[v]+curback<cap/2) dfs(v,cursend+cap/2-bike[v]-curback,0,curcost+cost[cur][v]);else dfs(v,cursend,curback+bike[v]-cap/2,curcost+cost[cur][v]);vis[v]=0;curpath.pop_back(); }}
// vis[cur]=0;curpath.pop_back();
}
int main(){scanf("%d %d %d %d",&cap,&n,&sp,&m);for(int i=1;i<=n;++i) scanf("%d",&bike[i]);for(int i=1,x,y,z;i<=m;++i){scanf("%d %d %d",&x,&y,&z);G[x].push_back(y);G[y].push_back(x);cost[x][y]=z;cost[y][x]=z; }dfs(0,0,0,0);printf("%d ",minsend);for(int i=0;i<anspath.size();++i)if(i==0) printf("%d",anspath[i]);else printf("->%d",anspath[i]); printf(" %d\n",minback);return 0;
} （更新：知道为什么了，，因为前面有的地方就return了，就没执行popback这行了，样例都过不了，所以保险起见，还是改成for循环中vis=1和vis=0吧，改成这样:）
void dfs(int cur,int cursend,int curback,int curcost){
// vis[cur]=1;curpath.push_back(cur);if(curcost>mincost) return ;if(cur==sp){if(curcost<mincost){mincost=curcost;minsend=cursend;minback=curback;anspath=curpath;return ;}if(curcost==mincost&&cursend<minsend){mincost=curcost;minsend=cursend;minback=curback;anspath=curpath;return ; }if(curcost==mincost&&cursend==minsend&&curback<minback){mincost=curcost;minsend=cursend;minback=curback;anspath=curpath;return ;}return ;}for(int i=0;i<G[cur].size();++i){int v=G[cur][i];if(!vis[v]){vis[v] = 1;curpath.push_back(v);if(bike[v]+curback<cap/2) dfs(v,cursend+cap/2-bike[v]-curback,0,curcost+cost[cur][v]);else dfs(v,cursend,curback+bike[v]-cap/2,curcost+cost[cur][v]);vis[v]=0;curpath.pop_back(); }}
// vis[cur]=0;curpath.pop_back();
}并且在主函数的dfs前：
vis[0]=1;curpath.push_back(0);
AC代码4：
但是感觉还是复杂度有点高，并且不稳定，我给他改了一下，变成简单易懂的了。。
#include<bits/stdc++.h>
using namespace std;
const int maxn=500+5,INF=0x3f3f3f3f;
int cap,n,sp,m,bike[maxn],cost[maxn][maxn];
vector<int> G[maxn],anspath,curpath;
int minsend=INF,minback=INF,mincost=INF,vis[maxn];
int dis[maxn];
void Dijkstra() {int all = n+1;bool visit[maxn] = {0};memset(dis,INF,sizeof dis);dis[0]=0;while(all--) {int minv,minw = INF;for(int i = 0; i<=n; i++) {if(visit[i]==0 && dis[i] < minw) minw=dis[i],minv=i;}visit[minv]=1;for(int i = 0; i<G[minv].size(); i++) {int now = G[minv][i];if(visit[now] == 0 && dis[now] > dis[minv] + cost[minv][now]) dis[now] = dis[minv] + cost[minv][now];}}
}
void dfs(int cur,int cursend,int curback,int curcost){if(curcost>mincost) return ;if(cur==sp){if(cursend<minsend){minsend=cursend;minback=curback;anspath=curpath;return ; }if(cursend==minsend&&curback<minback){minsend=cursend;minback=curback;anspath=curpath;return ;}return ;}for(int i=0;i<G[cur].size();i++){int v=G[cur][i];if(!vis[v]){vis[v] = 1;curpath.push_back(v);if(bike[v]+curback<cap/2) dfs(v,cursend+cap/2-bike[v]-curback,0,curcost+cost[cur][v]);else dfs(v,cursend,curback+bike[v]-cap/2,curcost+cost[cur][v]);vis[v]=0;curpath.pop_back();}}
}
int main(){scanf("%d %d %d %d",&cap,&n,&sp,&m);for(int i=1;i<=n;++i) scanf("%d",&bike[i]);for(int i=1,x,y,z;i<=m;++i){scanf("%d %d %d",&x,&y,&z);G[x].push_back(y);G[y].push_back(x);cost[x][y]=z;cost[y][x]=z; }Dijkstra();mincost = dis[sp];vis[0]=1;curpath.push_back(0); dfs(0,0,0,0);printf("%d ",minsend);for(int i=0;i<anspath.size();++i)if(i==0) printf("%d",anspath[i]);else printf("->%d",anspath[i]); printf(" %d\n",minback);return 0;
}
AC代码5：
还算能看懂的：（其实和AC代码3差不多、、）
#include<iostream>
#include<fstream>
#include<vector>
using namespace std;
#define MAX 505
#define INF 10000int cap,N,sp,M;
int vex;
int dist[MAX][MAX];
int bike[MAX];
#define PF cap/2vector<int> curpath;
vector<int> shortpath;
int minsend=INF,minback=INF;
int minlen=INF;
int cursend=0,curback=0;
int curlen=0;
bool visit[MAX]= {0};void dfs(int cur) {if(curlen>minlen)return;if(cur==sp) {//到达目标点，看是否最优if(curlen<minlen) {minlen=curlen;minsend=cursend;minback=curback;shortpath=curpath;} else if(curlen==minlen) {if(cursend<minsend||(cursend==minsend&&curback<minback)) {minsend=cursend;minback=curback;shortpath=curpath;}}return;}for(int i=1; i<vex; i++) {if(visit[i]==true||dist[cur][i]==INF)continue;int lastsend=cursend;int lastback=curback;//计算到达当前点的send和back数if(bike[i]+curback<PF) {cursend+=PF-bike[i]-curback;curback=0;} else {curback=bike[i]+curback-PF;}visit[i]=true;curpath.push_back(i);curlen+=dist[cur][i];dfs(i);curpath.pop_back();visit[i]=false;curlen-=dist[cur][i];cursend=lastsend;curback=lastback;}
}
int main() {cin>>cap>>N>>sp>>M;//初始化,距离置为INFvex=N+1;for(int i=0; i<vex; i++) {for(int j=0; j<vex; j++) {dist[i][j]=dist[j][i]=INF;}}for(int i=1; i<vex; i++) cin>>bike[i];for(int k=0; k<M; k++) {int i,j;cin>>i>>j>>dist[i][j];dist[j][i]=dist[i][j];}dfs(0);printf("%d 0",minsend);for(int i=0; i<shortpath.size(); i++) {printf("->%d",shortpath[i]);}printf(" %d",minback);return 0;
}
ps：wjh大佬写了个Dijkstra不加dfs，有点小问题但是也AC了，但是这个代码也告诉我们为什么这题不能用Dijkstra而应该加上dfs。
#include<bits/stdc++.h>
using namespace std;
struct node {int id;int d;int next;
} side[12000];
int head[1200];
int dis[1200];
int book[1200];
int cm,n,m,sp,cmax;
int num[1000];
int f[1000];
int dc[1000],tz[1000];
int cnt=0;
void init() {for(int i=0; i<=800; i++) {f[i]=i;}memset(dc,0x3f3f3f3f,sizeof(dc));memset(tz,0,sizeof(tz));memset(num,0,sizeof(num));memset(head,-1,sizeof(head));memset(dis,0x3f3f3f3f,sizeof(dis));memset(book,0,sizeof(book));cnt=0;
}
void add(int x,int y,int d) {side[cnt].id=y;side[cnt].d=d;side[cnt].next=head[x];head[x]=cnt++;
}
struct nod {int id;int d;nod() {}nod(int id,int d):id(id),d(d) {}friend bool operator<(nod a,nod b) {return a.d>b.d;}
};
void printpath(int x) {if(f[x]==x) {printf("0");return;} else {printpath(f[x]);printf("->%d",x);}
}
void dijkstra(int sx,int ex) {priority_queue<nod> q;q.push(nod(sx,0));dis[sx]=0;dc[sx]=0;while(!q.empty()) {nod nn=q.top();q.pop();if(book[nn.id])continue;book[nn.id]=1;if(nn.id==ex)break;for(int i=head[nn.id]; i!=-1; i=side[i].next) {int ty=side[i].id;if(dis[ty]>dis[nn.id]+side[i].d) {dis[ty]=dis[nn.id]+side[i].d;f[ty]=nn.id;if(cm>=num[ty]+tz[nn.id]) {tz[ty]=0;dc[ty]=dc[nn.id]+cm-(num[ty]+tz[nn.id]);} else {dc[ty]=dc[nn.id];tz[ty]=tz[nn.id]+num[ty]-cm;}q.push(nod(ty,dis[ty]));} else if(dis[ty]==dis[nn.id]+side[i].d) {
// &&dc[ty]<if(cm>=num[ty]+tz[nn.id]) {if(dc[ty]>dc[nn.id]+cm-num[ty]+tz[nn.id]) {tz[ty]=0;dc[ty]=dc[nn.id]+cm-(num[ty]+tz[nn.id]);f[ty]=nn.id;q.push(nod(ty,dis[ty]));}} else {if(dc[ty]>dc[nn.id]) {dc[ty]=dc[nn.id];tz[ty]=tz[nn.id]+num[ty]-cm;f[ty]=nn.id;q.push(nod(ty,dis[ty]));}}} }}cout<<dc[ex]<<" ";printpath(ex);cout<<" "<<tz[ex]<<endl;
}int main() {scanf("%d%d%d%d",&cmax,&n,&sp,&m);cm=cmax/2;int x,y,w;init();//初始化for(int i=1; i<=n; i++) {scanf("%d",&num[i]);}for(int i=0; i<m; i++) {scanf("%d%d%d",&x,&y,&w);add(x,y,w);add(y,x,w);}dijkstra(0,sp);return 0;
}总结：
这题不能用Dijkstra啊，因为在贪心的过程中只是dis数组是有效信息（顶多依据题意加一个最短路条数），其他的还有很多信息都丢失了。比如这组样例，其中Cmax=10（也就是要凑5），第一个圈为PBMC，每个圈中的数字代表自行车数;
如果最后一个圈内数字是0，那么为①路径，如果圈内数字是5，那么为②路径。所以具体哪个路径还是要搜索去看，而不能DIjkstra就定终生。因为比如：在A这个点的时候，是①和②两条路径择优选择了，没问题，如果到此为止了肯定我们要②路径。但是问题就在于，对于后面的点，未必会继承这一特性啊！（以后自己造样例的时候也可以这样造，造一个样例来说明不满足“可继承”特性的）换句话说这题的条件和需要加的限制很多，并无法证明满足无后效性，所以会Dijkstra丢掉很多信息，而应该用dfs，搜索每一条路径，这样就不会有漏下的情况。
读到这里，我也是理解了前面那个“分析”中说的不能只用DIjkstra的原因。
