Distance Queries
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 12846 | Accepted: 4552 | |
| Case Time Limit: 1000MS | ||
Description
Farmer 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
Hint
Farms 2 and 6 are 20+3+13=36 apart.
题目连接：POJ 1986
简单模版题，一棵树中两点的距离$d(u,v)$可以用$d[u]+d[v]-2*d[lca(u,v)]$来求得，其中$d_i$是你设定的根到某一点$i$的距离，那显然首先随便找个点进行最短路或者直接DFS获得d数组，再Tarjan得出答案
代码：
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=40010;
struct edge
{
int to;
int nxt;
int w;
};
struct query
{
int to;
int nxt;
int id;
}; edge E[N<<1];
query Q[N<<1];
int head[N],rhead[N],tot,rtot;
int d[N],dx[N],vis[N],in[N];
int pre[N],ances[N]; void init()
{
CLR(head,-1);
CLR(rhead,-1);
tot=rtot=0;
CLR(d,0);
for (int i=0; i<N; ++i)
{
pre[i]=i;
ances[i]=0;
}
CLR(vis,0);
CLR(in,0);
CLR(dx,0);
}
int Find(int n)
{
if(pre[n]==n)
return n;
return pre[n]=Find(pre[n]);
}
inline void add(int s,int t,int d)
{
E[tot].to=t;
E[tot].w=d;
E[tot].nxt=head[s];
head[s]=tot++;
}
inline void addquery(int s,int t,int id)
{
Q[rtot].id=id;
Q[rtot].to=t;
Q[rtot].nxt=rhead[s];
rhead[s]=rtot++;
}
void LCA(int u)
{
vis[u]=1;
ances[u]=u;
int i,v;
for (i=head[u]; ~i; i = E[i].nxt)
{
v = E[i].to;
if(!vis[v])
{
LCA(v);
pre[v]=u;
ances[Find(u)]=u;
}
}
for (i=rhead[u]; ~i; i = Q[i].nxt)
{
v=Q[i].to;
if(vis[v])
dx[Q[i].id]=d[u]+d[v]-(d[ances[Find(v)]]<<1);
}
}
void dfs(int u,int fa,int sum)
{
d[u]=sum;
for (int i=head[u]; ~i; i = E[i].nxt)
{
int v=E[i].to;
if(v!=fa)
dfs(v,u,sum+E[i].w);
}
}
int main(void)
{
int n,m,a,b,c,i,k;
char nouse[5];
while (~scanf("%d%d",&n,&m))
{
init();
for (i=0; i<m; ++i)
{
scanf("%d%d%d%s",&a,&b,&c,nouse);
add(a,b,c);
add(b,a,c);
++in[b];
}
scanf("%d",&k);
for (i=0; i<k; ++i)
{
scanf("%d%d",&a,&b);
addquery(a,b,i);
addquery(b,a,i);
}
for (i=1; i<=n; ++i)
{
if(!in[i])
{
dfs(i,-1,0);
LCA(i);
break;
}
}
for (i=0; i<k; ++i)
printf("%d\n",dx[i]);
}
return 0;
}