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.
题解：
LCA问题,dis[x]表示x到根节点的距离,设x,y的最近公共祖先为z,则x,y之间的距离为dis[x]+dis[y]-2*dis[z]。
倍增：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn=1000+5; inline int read() { int x=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9') {if(ch=='-')f=-1; ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} return x*f; } int n,m,num; int head[maxn],f[maxn][20],dep[maxn],dis[maxn]; bool vis[maxn]; struct node { int next,to,dist; }e[maxn<<1]; inline void add(int from,int to,int dist) { e[++num].next=head[from]; e[num].to=to; e[num].dist=dist; head[from]=num; } inline void dfs(int x,int d) { vis[x]=1;dep[x]=d; for(int i=head[x];i;i=e[i].next) { int to=e[i].to; if(!vis[to]) { dis[to]=dis[x]+e[i].dist; f[to][0]=x; dfs(to,d+1); } } } inline int lca(int a,int b) { if(dep[a]<dep[b]){int t=a;a=b;b=t;} int d=dep[a]-dep[b]; for(int i=0;i<=10;i++) if(d&(1<<i)) a=f[a][i]; if(a==b) return a; for(int i=10;i>=0;i--) if(f[a][i]!=f[b][i]) { a=f[a][i]; b=f[b][i]; } return f[a][0]; } int main() { n=read();m=read(); for(int i=1;i<n;i++) { int x,y,z; x=read();y=read();z=read(); add(x,y,z);add(y,x,z); } dfs(1,1); for(int j=1;j<=10;j++) for(int i=1;i<=n;i++) f[i][j]=f[f[i][j-1]][j-1]; for(int i=1;i<=m;i++) { int a,b; a=read();b=read(); printf("%d ",dis[a]+dis[b]-(dis[lca(a,b)]<<1)); } return 0; }
tarjan算法：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn=1000+5; inline int read() { int x=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9') {if(ch=='-')f=-1; ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} return x*f; } int n,m,num,qnum; int head[maxn],qhead[maxn],father[maxn],a[maxn][3],dis[maxn]; bool vis[maxn]; struct node { int next,to,v; }e[maxn<<1],q[maxn<<1]; void add(int from,int to,int v) { e[++num].next=head[from]; e[num].to=to; e[num].v=v; head[from]=num; } void qadd(int from,int to,int v) { q[++qnum].next=qhead[from]; q[qnum].to=to; q[qnum].v=v; qhead[from]=qnum; } int find(int x) { if(x!=father[x]) father[x]=find(father[x]); return father[x]; } void merge(int x,int y) { int r1=find(x); int r2=find(y); father[r1]=r2; } void tarjan(int x) { vis[x]=1; for(int i=qhead[x];i;i=q[i].next) { int to=q[i].to,v=q[i].v; if(vis[to]) a[v][2]=find(to); } for(int i=head[x];i;i=e[i].next) { int to=e[i].to; if(!vis[to]) { dis[to]=dis[x]+e[i].v; tarjan(to); merge(to,x); } } } int main() { n=read();m=read(); for(int i=1;i<=n;i++) father[i]=i; for(int i=1;i<n;i++) { int x,y,z; x=read();y=read();z=read(); add(x,y,z);add(y,x,z); } for(int i=1;i<=m;i++) { a[i][0]=read();a[i][1]=read(); qadd(a[i][0],a[i][1],i); qadd(a[i][1],a[i][0],i); } dis[1]=0; tarjan(1); for(int i=1;i<=m;i++) printf("%d ",dis[a[i][0]]+dis[a[i][1]]-(dis[a[i][2]]<<1)); return 0; }