poj-1986 Distance Queries(lca+ST+dfs)

题目链接：Distance Queries

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 11531		Accepted: 4068
Case Time Limit: 1000MS

DescriptionFarmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this problem consists of the same input as in "Navigation Nightmare",followed by a line containing a single integer K, followed by K "distance queries". Each distance query is a line of input containing two integers, giving the numbers of two farms between which FJ is interested in computing distance (measured in the length of the roads along the path between the two farms). Please answer FJ's distance queries as quickly as possible!Input* Lines 1..1+M: Same format as "Navigation Nightmare"* Line 2+M: A single integer, K. 1 <= K <= 10,000* Lines 3+M..2+M+K: Each line corresponds to a distance query and contains the indices of two farms.Output* Lines 1..K: For each distance query, output on a single line an integer giving the appropriate distance.Sample Input

7 6
1 6 13 E
6 3 9 E
3 5 7 S
4 1 3 N
2 4 20 W
4 7 2 S
3
1 6
1 4
2 6

Sample Output

13
3
36

题意：

给一棵树;问两点之间的距离;

思路：

lca的模板题,一开始莫名其妙的wa了,看了好久才发现是一个地方写反了;后来又tle,把cin改成scanf就好了;看来还是不能用cin;

AC代码：

/*2014300227    1986    Accepted    13984K    485MS    G++    2077B*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int N=4e4+4;
typedef long long ll;
const double PI=acos(-1.0);
int n,m,cnt,head
,vis
,a[2*N],dep
,first
,dis
,num,dp[2*N][30],p
;

struct Edge
{
int to,next,val;
};
Edge edge[2*N];
void add_edge(int s,int e,int va)
{
edge[cnt].to=e;
edge[cnt].next=head[s];
edge[cnt].val=va;
head[s]=cnt++;
}
int dfs(int x,int deep)
{
vis[x]=1;
first[x]=num;
a[num++]=x;
dep[x]=deep;
for(int i=head[x];i!=-1;i=edge[i].next)
{
int y=edge[i].to;
if(!vis[y])
{
dis[y]=dis[x]+edge[i].val;
dfs(y,deep+1);
a[num++]=x;
}
}
}
int RMQ()
{
for(int i=0;i<num;i++)
{
dp[i][0]=a[i];
}
for(int j=1;(1<<j)<=num;j++)
{
for(int i=0;i+(1<<j)-1<num;i++)
{
if(dep[dp[i][j-1]]<dep[dp[i+(1<<(j-1))][j-1]])dp[i][j]=dp[i][j-1];
else dp[i][j]=dp[i+(1<<(j-1))][j-1];
}
}
}
int query(int l,int r)
{
int temp=(int)(log((r-l+1)*1.0)/log(2.0));
if(dep[dp[l][temp]]<dep[dp[r-(1<<temp)+1][temp]])return dp[l][temp];
else return dp[r-(1<<temp)+1][temp];
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
cnt=0;
num=0;
for(int i=0;i<=n;i++)
{
vis[i]=0;
head[i]=-1;
}
int u,v,w;
char s[10];
for(int i=0;i<m;i++)
{
scanf("%d%d%d%s",&u,&v,&w,s);
//cin>>u>>v>>w>>s;
add_edge(u,v,w);
add_edge(v,u,w);
}
dis[1]=0;
dfs(1,0);
RMQ();
int q,fx,fy;
scanf("%d",&q);
while(q--)
{
scanf("%d%d",&fx,&fy);
int lca;
if(first[fx]<first[fy])
{
lca=query(first[fx],first[fy]);
}
else lca=query(first[fy],first[fx]);
printf("%d\n",dis[fx]+dis[fy]-2*dis[lca]);
}
}
return 0;
}