[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
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
输入样例#1：
4 2 2 1 2 4 3 2 1 4 3 1 2 3 2
输出样例#1：
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.
树剖LCA模板
1 #include <cstdio>
2 #include <cctype>
3 #include <queue>
4
5 const int MAXN=1010;
6 const int INF=0x3f3f3f3f;
7
8 int n,q,inr;
9
10 int dep[MAXN],fa[MAXN],son[MAXN],siz[MAXN],id[MAXN],top[MAXN],dis[MAXN];
11
12 bool vis[MAXN];
13
14 struct node {
15 int to;
16 int next;
17 int val;
18 node() {};
19 node(int to,int val,int next):to(to),val(val),next(next) {}
20 };
21 node e[MAXN<<1];
22
23 int head[MAXN],tot;
24
25 inline void read(int&x) {
26 int f=1;register char c=getchar();
27 for(x=0;!isdigit(c);c==‘-‘&&(f=-1),c=getchar());
28 for(;isdigit(c);x=x*10+c-48,c=getchar());
29 x=x*f;
30 }
31
32 inline void add(int x,int y,int val) {
33 e[++tot]=node(y,val,head[x]);
34 head[x]=tot;
35 e[++tot]=node(x,val,head[y]);
36 head[y]=tot;
37 }
38
39 void SPFA() {
40 std::queue<int> q;
41 for(int i=1;i<=n;++i) vis[i]=false,dis[i]=INF,son[i]=-1;
42 dis[1]=0;
43 q.push(1);
44 while(!q.empty()) {
45 int u=q.front();
46 q.pop();
47 for(int i=head[u];i;i=e[i].next) {
48 int v=e[i].to;
49 if(dis[v]>dis[u]+e[i].val) {
50 dis[v]=dis[u]+e[i].val;
51 if(!vis[v]) q.push(v),vis[v]=true;
52 }
53 }
54 }
55 }
56
57 void DFS_1(int u,int f) {
58 dep[u]=dep[f]+1;
59 fa[u]=f;
60 siz[u]=1;
61 for(int i=head[u];i;i=e[i].next) {
62 int v=e[i].to;
63 if(v==f) continue;
64 DFS_1(v,u);
65 siz[u]+=siz[v];
66 if(son[u]==-1||siz[son[u]]<siz[v]) son[u]=v;
67 }
68 }
69
70 void DFS_2(int u,int tp) {
71 top[u]=tp;
72 id[u]=++inr;
73 if(son[u]==-1) return;
74 DFS_2(son[u],tp);
75 for(int i=head[u];i;i=e[i].next) {
76 int v=e[i].to;
77 if(v==fa[u]||v==son[u]) continue;
78 DFS_2(v,v);
79 }
80 }
81
82 inline int LCA(int x,int y) {
83 while(top[x]!=top[y]) {
84 if(dep[top[x]]<dep[top[y]]) x^=y^=x^=y;
85 x=fa[top[x]];
86 }
87 if(dep[x]>dep[y]) x^=y^=x^=y;
88 return x;
89 }
90
91 int hh() {
92 read(n);read(q);
93 for(int x,y,z,i=1;i<n;++i) {
94 read(x);read(y);read(z);
95 add(x,y,z);
96 }
97 SPFA();
98 DFS_1(1,0);
99 DFS_2(1,1);
100 for(int x,y,i=1;i<=q;++i) {
101 read(x);read(y);
102 int lca=LCA(x,y);
103 printf("%d\n",dis[x]+dis[y]-2*dis[lca]);
104 }
105 return 0;
106 }
107
108 int sb=hh();
109 int main(int argc,char**argv) {;}
代码