K. Another Shortest Path Problem
time limit per test
3.0 s
memory limit per test
256 MB
input
standard input
output
standard output
Shortest path problems have always been a part of competitive programming, appearing in many contests all around the world. In order to keep that tradition, we created another one:
You are given an undirected connected weighted graph with N nodes and N edges, and you have to answer Q queries. Each query consists of two nodes X and Y, and you have to find the length of the shortest path between nodes X and Y.
Input
The first line contains a single integer T, the number of test cases.
Each test case starts with a line containing two integers N and Q, the number of nodes (and edges) and the number of queries. (3 ≤ N ≤ 105) (1 ≤ Q ≤ 105)
Each of the following N lines contain the description of the edges. The ith line contains 3 space-separated integers ui, vi, and wi. This means that there is an undirected edge between nodes ui and vi, with a weight of wi. (1 ≤ ui, vi ≤ N) (1 ≤ wi ≤ 105)
Then Q lines follow, the ith line contains two integers X and Y. This means that you need to find the shortest path between nodes X and Y. (1 ≤ X, Y ≤ N)
It is guaranteed that the graph contains no self loops or multiple edges.
Output
For each test case, and for each query, print one line containing one integer, the shortest path between X and Y.
Example
input
Copy
1 6 3 1 2 2 1 3 4 2 6 3 3 4 1 3 5 10 3 6 6 1 4 2 5 3 2
output
Copy
5 16 6
题意：给你一个含n条边的带权无向连通图，q次查询，每次查询两点间的最短距离。
思路：LCA+思维。
设a，b两点间的距离为f(a,b) 则f(a,b)=dis[a]+dis[b]-2*dis[lca(a,b)];
由于n条边，因此我们先任取一条边，设这条边为X，Y，权值为Z，设查询的点为x，y，则答案为
min(f(a,b),f(a,X)+f(b,X),f(a,Y)+f(b,Y),f(a,X)+f(b,Y)+Z,f(a,Y)+f(b,X)+Z);
注意long long。这道题是在做完hdu5296后突然想起来的，其实这道题就是简单的lca应用，之前居然没做出来。。。
代码：
#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=400010;
int n,m,q,tot,tot2,cnt,tmp,ans;
int f[maxn][20];
int head[maxn];
ll dis[maxn];
int dfn[maxn],d[maxn];
int dep[maxn],e[maxn],pos[maxn];
bool vis[maxn];
struct node
{
int to,nex;
ll w;
}a[maxn];
void add(int u,int v,ll w)
{
a[cnt].to=v;
a[cnt].w=w;
a[cnt].nex=head[u];
head[u]=cnt++;
}
void init()
{
cnt=tot=tot2=0;
memset(head,-1,sizeof(head));
memset(pos,-1,sizeof(pos));
memset(vis,0,sizeof(vis));
dis[1]=0;
}
void dfs(int u,int deep)//1
{
if(pos[u]!=-1)return;
dfn[tot2]=u;d[u]=tot2++;
pos[u]=tot;e[tot]=u;dep[tot++]=deep;
for(int i=head[u];i!=-1;i=a[i].nex)
{
int v=a[i].to;
if(pos[v]==-1)
{
dis[v]=dis[u]+a[i].w;
dfs(v,deep+1);
e[tot]=u;dep[tot++]=deep;
}
}
}
void rmq(int n)//2
{
for(int i=1;i<=n;i++)f[i][0]=i;
for(int j=1;(1<<j)<=n;j++)
for(int i=1;i+(1<<j)-1<=n;i++)
{
if(dep[f[i][j-1]]<dep[f[i+(1<<(j-1))][j-1]]) f[i][j]=f[i][j-1];
else f[i][j]=f[i+(1<<(j-1))][j-1];
}
}
int RMQ(int l,int r)
{
int k=(int)(log((double)(r-l+1))/log(2.0));
if(dep[f[l][k]]<dep[f[r-(1<<k)+1][k]]) return f[l][k];
else return f[r-(1<<k)+1][k];
}
int lca(int x,int y)//3
{
if(pos[x]<pos[y]) return e[RMQ(pos[x],pos[y])];
else return e[RMQ(pos[y],pos[x])];
}
ll cal(int x,int y)
{
return dis[x]+dis[y]-2*dis[lca(x,y)];
}
int main()
{
int T,cas=1;
scanf("%d",&T);
while(T--)
{
init();
scanf("%d %d",&n,&q);
for(int i=0;i<n-1;i++)
{
int x,y;ll z;
scanf("%d%d%lld",&x,&y,&z);
add(x,y,z);
add(y,x,z);
}
int X,Y;ll Z;
scanf("%d%d%lld",&X,&Y,&Z);
//printf("Case #%d:\n",cas++);
dfs(1,0);
rmq(2*n-1);//4
ll ans=0;
while(q--)
{
int u,v;
scanf("%d %d",&u,&v);
ans=cal(u,v);
ans=min(ans,cal(u,X)+cal(v,X));
ans=min(ans,cal(u,Y)+cal(v,Y));
ans=min(ans,cal(u,X)+cal(v,Y)+Z);
ans=min(ans,cal(u,Y)+cal(v,X)+Z);
printf("%lld\n",ans);
}
}
return 0;
}