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

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复制

4 2 2 1 2 4 3 2 1 4 3 1 2 3 2

输出 #1复制

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
      
       #include
       
        #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;}