[USACO08OCT]牧场散步Pasture Walking

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

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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 #1

输出 #1

2 
7

说明/提示

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.

Floyd 的复杂度大家都知道是 $O(n^3)O(n3)，但是我们可以剪枝。我们都知道 Floyd 初始需要把 distdist 数组初始化为 \infty∞，所以当我们进行第二层循环时，如果当前 dist(i, k) = \inftydist(i,k)=∞ 就可以直接跳过。$

#include<cstdio>
#include<cstring>
#define min(a, b) ((a) < (b) ? (a) : (b))

using namespace std;
const int MaxN = 1e3;
const int INF = 0x3f3f3f3f;
int d[MaxN + 1][MaxN + 1];
int N, Q;

inline int read(){
    int s=0,w=1;
    char ch=getchar();
    while(ch<'0'||ch>'9'){
        if(ch=='-'){
            w=-1;
        }
        ch=getchar();
    }
    while(ch>='0'&&ch<='9'){
        s=s*10+ch-'0';
        ch=getchar();
    }
    return s*w;
}

int main(){
    memset(d, 0x3f, sizeof(d));
    N=read();
    Q=read();
    for (int i = 1, A, B, L; i < N; ++i) {
        A=read();
        B=read();
        L=read();
        d[A][B] = d[B][A] = L;
    }
    for (int k = 1; k <= N; ++k)
        for (int i = 1; i <= N; ++i) {
            if (d[i][k] == INF){
                continue;
            }
            for (int j = 1; j <= N; ++j){
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
            }
        }
    for (int P1, P2; Q--; ) {
        P1=read();
        P2=read();
        printf("%d\n", d[P1][P2]);
    }
    return 0;
}