题目描述
The N cows (2 <= N <= 1,000) conveniently numbered 1..N are grazing among the N pastures also conveniently numbered 1..N. Most conveniently of all, cow i is grazing in pasture i.
Some pairs of pastures are connected by one of N-1 bidirectional walkways that the cows can traverse. Walkway i connects pastures A_i and B_i (1 <= A_i <= N; 1 <= B_i <= N) and has a length of L_i (1 <= L_i <= 10,000).
The walkways are set up in such a way that between any two distinct pastures, there is exactly one path of walkways that travels between them. Thus, the walkways form a tree.
The cows are very social and wish to visit each other often. Ever in a hurry, they want you to help them schedule their visits by computing the lengths of the paths between 1 <= L_i <= 10,000 pairs of pastures (each pair given as a query p1,p2 (1 <= p1 <= N; 1 <= p2 <= N).
POINTS: 200
有N(2<=N<=1000)头奶牛，编号为1到W，它们正在同样编号为1到N的牧场上行走.为了方 便，我们假设编号为i的牛恰好在第i号牧场上.
有一些牧场间每两个牧场用一条双向道路相连，道路总共有N - 1条，奶牛可以在这些道路 上行走.第i条道路把第Ai个牧场和第Bi个牧场连了起来(1 <= A_i <= N; 1 <= B_i <= N)，而它的长度 是 1 <= L_i <= 10,000.在任意两个牧场间，有且仅有一条由若干道路组成的路径相连.也就是说，所有的道路构成了一棵树.
奶牛们十分希望经常互相见面.它们十分着急，所以希望你帮助它们计划它们的行程，你只 需要计算出Q(1 < Q < 1000)对点之间的路径长度•每对点以一个询问p1,p2 (1 <= p1 <= N; 1 <= p2 <= N). 的形式给出.
输入输出格式
输入格式：* Line 1: Two space-separated integers: N and Q
* Lines 2..N: Line i+1 contains three space-separated integers: A_i, B_i, and L_i
* Lines N+1..N+Q: Each line contains two space-separated integers representing two distinct pastures between which the cows wish to travel: p1 and p2
输出格式：* Lines 1..Q: Line i contains the length of the path between the two pastures in query i.
输入输出样例
说明
Query 1: The walkway between pastures 1 and 2 has length 2.
Query 2: Travel through the walkway between pastures 3 and 4, then the one between 4 and 1, and finally the one between 1 and 2, for a total length of 7.
LCA+bfs 即可；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}
int edge[maxn], ver[maxn], head[maxn], nxt[maxn];
int dp[maxn][25];
int dep[maxn];
int dis[maxn];
int t, cnt;
queue<int>q;
void addedge(int x, int y, int w) {
ver[++cnt] = y; edge[cnt] = w; nxt[cnt] = head[x]; head[x] = cnt;
}
void bfs() {
q.push(1); dis[1] = 0; dep[1] = 1;
while (!q.empty()) {
int x = q.front(); q.pop();
for (int i = head[x]; i; i = nxt[i]) {
int y = ver[i];
if (dep[y])continue;
dep[y] = dep[x] + 1;
dis[y] = dis[x] + edge[i]; dp[y][0] = x;
for (int j = 1; j <= t; j++) {
dp[y][j] = dp[dp[y][j - 1]][j - 1];
}
q.push(y);
}
}
}
int lca(int x, int y) {
if (dep[x] > dep[y])swap(x, y);
for (int i = t; i >= 0; i--) {
if (dep[dp[y][i]] >= dep[x])y = dp[y][i];
}
if (x == y)return x;
for (int i = t; i >= 0; i--) {
if (dp[y][i] != dp[x][i])y = dp[y][i], x = dp[x][i];
}
return dp[x][0];
}
int main()
{
//ios::sync_with_stdio(0);
int n;
rdint(n); t = 20;
int q; rdint(q);
for (int i = 1; i < n; i++) {
int x, y; rdint(x); rdint(y); int w; rdint(w);
addedge(x, y, w); addedge(y, x, w);
}
bfs();
while (q--) {
int a, b; rdint(a); rdint(b);
cout << dis[a] + dis[b] - 2 * dis[lca(a, b)] << endl;
}
return 0;
}