[USACO08OCT]牧场散步Pasture Walking BZOJ1602 LCA

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

输入样例#1：
复制

输出样例#1： 复制

2

7

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;

}