Problem Description
There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always
unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.
unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.
Input
First line is a single integer T(T<=10), indicating the number of test cases.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road
connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road
connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
Output
For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.
Sample Input
2 3 2 1 2 10 3 1 15 1 2 2 3 2 2 1 2 100 1 2 2 1
Sample Output
10 25 100 100
Tarjan算法：
/*//最近公共祖先
题目描述：给定一棵带权树，求u，v两节点的最短长度
算法思想：找到u，v两点的最近公共祖先lca(v)
若dis[u]表示根节点到u的长度，则res=dis[u]+dis[v]-2*dis[lca(v)]
*/
#include <iostream>
#include <string>
#define NN 40002 // number of house
#define MM 202 // number of query
using namespace std;
typedef struct node{
int v;//节点
int d;//v节点与nxt相连的权值
struct node *nxt;
}NODE;
NODE *Link1[NN];//邻接表，Link1[u]表示与u相连所有节点链
NODE edg1[NN * 2]; // 树中的边
NODE *Link2[NN];
NODE edg2[NN * 2]; // 询问的点对
int idx1, idx2, N, M;
int res[MM][3]; // 记录结果，res[i][0]: u res[i][1]: v res[i][2]: lca(u, v)
int fat[NN];//并查集表
int vis[NN];//记录访问点
int dis[NN];//根到i节点的长度
//前插法
void Add(int u, int v, int d, NODE edg[], NODE *Link[], int &idx){
edg[idx].v = v;
edg[idx].d = d;
edg[idx].nxt = Link[u];
Link[u] = edg + idx++;
edg[idx].v = u;
edg[idx].d = d;
edg[idx].nxt = Link[v];
Link[v] = edg + idx++;
}
int find(int x){ // 并查集路径压缩
if(x != fat[x]){
return fat[x] = find(fat[x]);
}
return x;
}
void Tarjan(int u){
vis[u] = 1;
fat[u] = u;
for (NODE *p = Link2[u]; p; p = p->nxt){
if(vis[p->v]){
res[p->d][2] = find(p->v); // 存的是最近公共祖先结点
}
}
for (NODE *p = Link1[u]; p; p = p->nxt){
if(!vis[p->v]){
dis[p->v] = dis[u] + p->d;
Tarjan(p->v);
fat[p->v] = u;
}
}
}
int main() {
freopen("C:\\in.txt","r",stdin);
int T, i, u, v, d;
scanf("%d", &T);
while(T--){
scanf("%d%d", &N, &M);
idx1 = 0;
memset(Link1, 0, sizeof(Link1));
for (i = 1; i < N; i++){
scanf("%d%d%d", &u, &v, &d);
Add(u, v, d, edg1, Link1, idx1);
}
idx2 = 0;
memset(Link2, 0, sizeof(Link2));
for (i = 1; i <= M; i++){
scanf("%d%d", &u, &v);
Add(u, v, i, edg2, Link2, idx2);
res[i][0] = u;
res[i][1] = v;
}
memset(vis, 0, sizeof(vis));
dis[1] = 0;
Tarjan(1);
for (i = 1; i <= M; i++){
printf("%d\n", dis[res[i][0]] + dis[res[i][1]] - 2 * dis[res[i][2]]);
}
}
return 0;
}