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.
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
纯倍增模板
顾名思义:就是利用倍增的思想来巧妙求LCA。
设grand[i][j]为节点i的2^j个祖先的编号,
pw[i][j]表示节点i到他的祖先2^i次方的距离 。
#include<stdio.h> #include<string.h> #include<vector> #include<stdlib.h> #include<math.h> #define max 40001 #define maxl 25 using namespace std; typedef struct { int from,to,w; }edge; vector<edge>edges; vector<int> G[max]; //保存边的数组 int grand[max][maxl], gw[max][maxl];//为节点i的2^j个祖先的编号, i到他上面祖先2^i次方的距离 int depth[max]; int n,m,N; void addedge(int u, int v, int w) { edge a = {u, v, w}, b = {v, u, w}; edges.push_back(a); edges.push_back(b); G[u].push_back(edges.size()-2); G[v].push_back(edges.size()-1); } void dfs(int x) { for(int i = 1; i <= N; i++) { grand[x][i] = grand[grand[x][i-1]][i-1]; gw[x][i] = gw[x][i-1] + gw[grand[x][i-1]][i-1]; } for(int i = 0; i < G[x].size(); i++) { edge e = edges[G[x][i]]; // G[x][i]表示坐标点 if(e.to != grand[x][0]) { depth[e.to] = depth[x] + 1; grand[e.to][0] = x; //与e.to相连那个节点的父亲等于x gw[e.to][0] = e.w; //与x相连那个节点的距离等于这条边的距离 dfs(e.to); //深搜往下面建 } } } void Init() { N = floor(log(n + 0.0) / log(2.0)); //最多能跳的2^i祖先, log2 n ; depth[1] = 0; memset(grand, 0, sizeof(grand)); memset(gw, 0, sizeof(gw)); dfs(1);//以1为根节点建树 } int lca(int a, int b) { if(depth[a] > depth[b]) swap(a, b); int ans = 0; for(int i = N; i >= 0; i--) { if(depth[a] < depth[b] && depth[a] <= depth[grand[b][i]]) { ans += gw[b][i]; b = grand[b][i];//先把深度较大的b往上跳 } } for(int j = N; j >= 0; j--)//在同一高度了,他们一起向上跳,跳他们不相同节点,当全都跳完之后grand[a][0]就是lca { if(grand[a][j] != grand[b][j]) { ans += gw[a][j]; ans += gw[b][j]; a = grand[a][j]; b = grand[b][j]; } } if(a != b) { ans += gw[a][0]; ans += gw[b][0]; } return ans; } int main() { int t ; scanf("%d", &t); while(t--) { scanf("%d%d", &n, &m); for(int i = 1; i < n; i++) { int x, y, w; scanf("%d%d%d", &x, &y, &w); addedge(x, y, w); } Init(); for(int i = 1; i <= m; i++) { int x,y; scanf("%d%d", &x, &y); printf("%d\n", lca(x,y)); } } return 0; }