P2912 [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.
输入输出格式
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
- Lines 1..Q: Line i contains the length of the path between the two pastures in query i.
输入输出样例
4 2 2 1 2 4 3 2 1 4 3 1 2 3 2
2 7
说明
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.
裸LCA,预处理的时候顺带求到根节点的路径长，答案为len[u] + len[v] - 2 * len[lca(va, vb）]
1 #include <bits/stdc++.h> 2 3 const int MAXN = 5000 + 10; 4 5 inline void read(int &x) 6 { 7 x = 0;char ch = getchar();char c = ch; 8 while(ch > '9' || ch < '0')c = ch, ch = getchar(); 9 while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar(); 10 if(c == '-')x = -x; 11 } 12 13 inline void swap(int &a, int &b) 14 { 15 int tmp = a; 16 a = b; 17 b = tmp; 18 } 19 20 struct Edge{int u,v,w,next;}edge[MAXN >> 1]; 21 int head[MAXN],root,cnt,deep[MAXN],len[MAXN],p[MAXN][20]; 22 int n,m,tmp1,tmp2,tmp3; 23 bool b[MAXN]; 24 25 inline void insert(int a, int b, int c) 26 { 27 edge[++cnt] = {a,b,c,head[a]}; 28 head[a] = cnt; 29 } 30 31 void dfs(int u, int step) 32 { 33 for(int pos = head[u];pos;pos = edge[pos].next) 34 { 35 int v = edge[pos].v; 36 if(!b[v]) 37 { 38 b[v] = true; 39 deep[v] = step + 1; 40 len[v] = len[u] + edge[pos].w; 41 p[v][0] = u; 42 dfs(v, step + 1); 43 } 44 } 45 } 46 47 inline void yuchuli() 48 { 49 b[root] = true; 50 len[root] = 0; 51 deep[root] = 0; 52 dfs(root, 0); 53 for(int i = 1;(1 << i) <= n;i ++) 54 { 55 for(int j = 1;j <= n;j ++) 56 { 57 p[j][i] = p[p[j][i - 1]][i - 1]; 58 } 59 } 60 } 61 62 inline int lca(int va, int vb) 63 { 64 if(deep[va] < deep[vb])swap(va, vb); 65 int M = 0; 66 for(;(1 << M) <= n;M ++); 67 M --; 68 for(int i = M;i >= 0;i --) 69 if(deep[va] - (1 << i) >= deep[vb]) 70 va = p[va][i]; 71 if(va == vb)return va; 72 for(int i = M;i >= 0;i --) 73 if(p[va][i] != p[vb][i]) 74 { 75 va = p[va][i]; 76 vb = p[vb][i]; 77 } 78 return p[va][0]; 79 } 80 81 int main() 82 { 83 read(n);read(m); 84 for(int i = 1;i < n;i ++) 85 { 86 read(tmp1);read(tmp2);read(tmp3); 87 insert(tmp1, tmp2, tmp3); 88 insert(tmp2, tmp1, tmp3); 89 } 90 root = 1; 91 yuchuli(); 92 for(int i = 1;i <= m;i ++) 93 { 94 read(tmp1);read(tmp2); 95 printf("%d ", len[tmp1] + len[tmp2] - 2 * len[lca(tmp1, tmp2)]); 96 } 97 return 0; 98 }