AtCoder Beginner Contest 070 Transit Tree Path（一道鸡贼的最短路径题）

D - Transit Tree Path
Time limit : 2sec / Memory limit : 256MB


Score : 400 points


Problem Statement
You are given a tree with N vertices.
Here, a tree is a kind of graph, and more specifically, a connected undirected graph with N−1 edges, where N is the number of its vertices.
The i-th edge (1≤i≤N−1) connects Vertices ai and bi, and has a length of ci.


You are also given Q queries and an integer K. In the j-th query (1≤j≤Q):


find the length of the shortest path from Vertex xj and Vertex yj via Vertex K.
Constraints
3≤N≤105
1≤ai,bi≤N(1≤i≤N−1)
1≤ci≤109(1≤i≤N−1)
The given graph is a tree.
1≤Q≤105
1≤K≤N
1≤xj,yj≤N(1≤j≤Q)
xj≠yj(1≤j≤Q)
xj≠K,yj≠K(1≤j≤Q)
Input
Input is given from Standard Input in the following format:


N  
a1 b1 c1  
:  
aN−1 bN−1 cN−1
Q K
x1 y1
:  
xQ yQ
Output
Print the responses to the queries in Q lines.
In the j-th line j(1≤j≤Q), print the response to the j-th query.

如题，最短路径，只是是必经过点到任一点的最短路径；

不多bb，代码如下：

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#include <map>#include <vector>#include <queue>using namespace std;const int N=1e5+10,INF=0x3f3f3f3f;typedef pair<int,int>P; typedef vector<int>::iterator ive;long long step[N];//步数，这里是权重 int k;map<int,vector<int> >m;map<P,int>mmp;/* P是点 ，int是亮点之间的权值*/void bfs()//以k为始，队列存点，遍历全部，广搜深搜都可以{    queue<int>q;    q.push(k);    step[k]=0;    while(!q.empty()){        int now=q.front();        q.pop();        for(ive it=m[now].begin();it!=m[now].end();it++){            if(step[now]+mmp[P(max(*it,now),min(*it,now))]<step[*it])//条件：预防以前的路再走一遍 ，如果有更省权重的方案，就更新 {                step[*it]=step[now]+mmp[P(max(*it,now),min(*it,now))];                q.push(*it);            }          }    }}int main(){    int n,a,b,c,q,x,y;    while(~scanf("%d",&n)){        memset(step,INF,sizeof(step));        m.clear();        mmp.clear();        n--;        while(n--){            scanf("%d%d%d",&a,&b,&c);            m[a].push_back(b);//存路径 ，因为是互通的，所以正反都存             m[b].push_back(a);            mmp[P(max(a,b),min(a,b))]=c;//存权重，节省空间，规定大的在前小的在后         }        scanf("%d%d",&q,&k);        bfs();        while(q--){            scanf("%d%d",&x,&y);            printf("%lld\n",step[x]+step[y]);        }    }    return 0;}