[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
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.
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 ",dis[x]+dis[y]-2*dis[lca]); 104 } 105 return 0; 106 } 107 108 int sb=hh(); 109 int main(int argc,char**argv) {;}