[USACO08OCT] pasture walks Pasture Walking

Title Description

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

There are N (2 <= N <= 1000) cows, numbered 1 to W, they are similarly numbered 1 to N pasture walking. For convenience, we assume that the number i of cattle just on the i-th pastures .

Some pasture between each two pastures are connected by a two-way road, the road has a total of N - 1 bar, cows can walk on these roads of the i th path Ai and Bi pastures up even pastures (1 <. = A_i <= N; 1 <= B_i <= N), and its length is 1 <= L_i <= 10,000 between any two pastures, and only one path is connected by a plurality of road is composed. He said that all roads constitute a tree.

The cows very often want to meet each other. They very anxious, so I hope you help them plan their trip, you only need to calculate the Q (1 <Q <1000) • The length of the path between each pair of points to a point of asking p1 , p2 (1 <= p1 <= N; 1 <= p2 <= N). given in the form.

Input Format

* 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

Output Format

* Lines 1..Q: Line i contains the length of the path between the two pastures in query i.

Sample input and output

Input # 1

Output # 1

2 
7

Description / Tips

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.

The complexity of the Floyd we all know is $n-^ 3) O ( the n- 3 ), but we can prune. We need to know the original Floyd dist D I S T array is initialized to \ infty ∞, so the second layer when we cycle, if the current dist (I, K) = \ infty D I S T ( I , K ) = ∞ can skip.$

#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;
}