HDU 2586 How far away ?

时间:2024-08-07 15:37:56

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18652    Accepted Submission(s):
7268

Problem Description
There are n houses in the village and some
bidirectional roads connecting them. Every day peole always like to ask like
this "How far is it if I want to go from house A to house B"? Usually it hard to
answer. But luckily int this village the answer is always unique, since the
roads are built in the way that there is a unique simple path("simple" means you
can't visit a place twice) between every two houses. Yout task is to answer all
these curious people.
Input
First line is a single integer T(T<=10), indicating
the number of test cases.
  For each test case,in the first line there are
two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses
and the number of queries. The following n-1 lines each consisting three numbers
i,j,k, separated bu a single space, meaning that there is a road connecting
house i and house j,with length k(0<k<=40000).The houses are labeled from
1 to n.
  Next m lines each has distinct integers i and j, you areato answer
the distance between house i and house j.
Output
For each test case,output m lines. Each line represents
the answer of the query. Output a bland line after each test case.
Sample Input
2
3 2
1 2 10
3 1 15
1 2
2 3

2 2
1 2 100
1 2
2 1

Sample Output
10
25
100
100
Source
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带权的LCA问题
我们用g[i]表示i号节点走到根的权值
那么两个点之间的路径权值为,$g[x]+g[y]-2*g[LCA(x,y)]$
大概是这个样子
被圆圈出来的是需要减去的
HDU 2586 How far away ?
 #include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=1e5+;
inline int read()
{
char c=getchar();int x=,f=;
while(c<''||c>'') {if(c=='-') f=-;c=getchar();}
while(c>=''&&c<='') x=x*+c-,c=getchar();return x*f;
}
int n,m,S=;
int f[MAXN][],deep[MAXN],g[MAXN];
struct node
{
int u,v,w,nxt;
}edge[MAXN];
int head[MAXN];
int num=;
inline void add_edge(int x,int y,int z)
{
edge[num].u=x;
edge[num].v=y;
edge[num].w=z;
edge[num].nxt=head[x];
head[x]=num++;
}
void dfs(int now)
{
for(int i=head[now];i!=-;i=edge[i].nxt)
if(deep[edge[i].v]==)
{
deep[edge[i].v]=deep[now]+;
f[edge[i].v][]=now;
g[edge[i].v]=g[now]+edge[i].w;
dfs(edge[i].v);
} }
inline void pre()
{
for(int i=;i<=;i++)
for(int j=;j<=n;j++)
f[j][i]=f[f[j][i-]][i-];
}
inline int LCA(int x,int y)
{
if(deep[x]<deep[y]) swap(x,y);
for(int i=;i>=;i--)
if(deep[f[x][i]]>=deep[y])
x=f[x][i];
if(x==y) return x; for(int i=;i>=;i--)
if(f[x][i]!=f[y][i])
x=f[x][i],y=f[y][i];
return f[x][];
}
int main()
{
int T=read();
while(T--)
{
n=read();m=read();
memset(head,-,sizeof(head));num=;
memset(f,,sizeof(f));
memset(deep,,sizeof(deep));
for(int i=;i<=n-;i++)
{
int x=read(),y=read(),z=read();
add_edge(x,y,z);
add_edge(y,x,z);
}
deep[S]=;
dfs(S);pre();
while(m--)
{
int x=read(),y=read();
printf("%d\n",g[x]+g[y]-*g[LCA(x,y)]);
}
} return ;
}