Codeforces Detour（Dijkstra +爆搜）
Description
After last year’s edition of the BAPC, you are still stuck in Delft. In order to participate again this year, you are going to Amsterdam by bus. During the journey you look out of the window and look for traffic signs that point in the direction of Amsterdam. To your surprise, you notice that the bus is never taking the roads that are pointed out by the signs! You think that the bus company might have chosen a route such that, at no intersection, the bus goes in the direction that is pointed to by the signs. Your friends, however, find this very unbelievable, and don’t think this is possible. Can you figure out whether there exists a bus route that satisfies this requirement? Note that a bus route never visits the same place twice. A traffic sign pointing in the direction of the shortest route to Amsterdam is placed at every intersection. You may assume that the input graph is both simple and connected, and that there is a unique optimal route to Amsterdam from every intersection.
Input
• A single line containing two integers: n (2 ≤ n ≤ 105 ), the number of intersections, and m (1 ≤ m ≤ 106 ), the number of undirected roads that connect the intersections. The intersections are numbered from 0 to n − 1. Delft is denoted by intersection i = 0 and Amsterdam is denoted by intersection i = 1.
• m lines that specify the roads – A road is specified by three integers, ai , bi (0 ≤ ai , bi < n and ai 6= bi) and di (0 ≤ di ≤ 500000), where ai and bi are ids of the two intersections that are connected by this road and di is the distance that the bus has to travel to get from ai to bi or vice versa.
Output
As output, give one of the following:
• A path from Delft to Amsterdam that satisfies the requirements, in case such a path exists. – A path is specified by a single line containing an integer k, the length of the path, followed by k integers pi that specify the intersections along the path in the order in which they are crossed, with p0 = 0 and pk−1 = 1.
• The text “impossible”, if such a path does not exist.
Sample
Sample Input 1
4 5
0 2 5
2 1 5
0 3 10
3 1 20
3 2 5
Sample Output 1
3 0 3 1
Sample Input 2
4 3
0 1 10
1 2 20
2 3 30
Sample Output 2
impossible
Sample Input 3
7 9
0 1 800
6 2 300
2 3 75
3 4 80
4 5 50
4 1 100
1 6 35
0 2 120
0 3 100
Sample Output 3
2 0 1
题目描述
现有n个路口，每个路口的路牌都会标记到 1 路口的最短路的下一个节点，现判断从0->1且不能走路牌上标记的方向，是否存在一种答案. 若存在输出路径，否则输出”impossible”.
解题思路
标记源点为1，先用dijkstar标记每个路口的路牌上指明的节点，然后标记的路线不可行，直接dfs爆搜一下标记可行路径.
代码实现
#includeque;
while(!que.empty())que.pop();
dist[start]=0;
que.push(qnode(start,0));
qnode tmp;
while(!que.empty())
{
tmp=que.top();
que.pop();
int u=tmp.v;
if(vis[u])continue;
vis[u]=true;
for(int i=head[u];i!=-1;i=edge[i].ne)
{
int v=edge[i].v;
int cost=edge[i].c;
if(!vis[v] && dist[v]>dist[u]+cost)
{
pre[v] = u;
dist[v] = dist[u]+cost;
que.push(qnode(v,dist[v]));
}
}
}
}
void print_ans(int point)
{
if(point==1)
{
cout<<" "<<0;
return ;
}
print_ans(point-1);
cout<<" "<1;
}
void dfs(int tmp,int pree)
{
ans[++len]=tmp;
vis[tmp]=1;
if(tmp==2)
{
flag=true;
cout<cout<return ;
}
for(int i=head[tmp];i!=-1;i=edge[i].ne)
{
int v=edge[i].v;
if(pre[tmp]==v || vis[v]) continue;
dfs(v,tmp);
if(flag) return ;
}
--len;
}
int main()
{
IO;
init();
int n,m;
int u,v,c;
cin>>n>>m;
for(int i=0;icin>>u>>v>>c;
addedge(u+1,v+1,c);
addedge(v+1,u+1,c);
}
Dijkstra(n,2);
memset(vis,0,sizeof(vis));
dfs(1,-1);
if(!flag) cout<<"impossible"<return 0;
}
Ps：Dijkstar 优化算法模板（from kuangbin）
/*
* 使用优先队列优化Dijkstra算法
* 复杂度O(ElogE)
* 注意对vectorE[MAXN]进行初始化后加边
*/
const int INF=0x3f3f3f3f;
const int MAXN=1000010;
struct qnode
{
int v;
int c;
qnode(int _v=0,int _c=0):v(_v),c(_c) {}
bool operator <(const qnode &r)const
{
return c>r.c;
}
};
struct Edge
{
int v,cost;
Edge(int _v=0,int _cost=0):v(_v),cost(_cost) {}
};
vectorque;
while(!que.empty())que.pop();
dist[start]=0;
que.push(qnode(start,0));
qnode tmp;
while(!que.empty())
{
tmp=que.top();
que.pop();
int u=tmp.v;
if(vis[u])continue;
vis[u]=true;
for(int i=0; iint v=E[tmp.v][i].v;
int cost=E[u][i].cost;
if(!vis[v]&&dist[v]>dist[u]+cost)
{
dist[v]=dist[u]+cost;
que.push(qnode(v,dist[v]));
}
}
}
}
void addedge(int u,int v,int w)
{
E[u].push_back(Edge(v,w));
}