[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
4 2 
2 1 2 
4 3 2 
1 4 3 
1 2 
3 2 
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;
}

 

posted @ 2018-11-23 23:30  NKDEWSM  阅读(111)  评论(0编辑  收藏  举报