hdu 2586 LCA模板題(離線和線上兩種解法)
阿新 • • 發佈:2019-01-24
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.
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
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.
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
/** hdu 2586 LCA模板題(離線演算法) 題目大意:給一個無根樹,有q個詢問,每個詢問兩個點,問兩點的距離。 解題思路:求出 lca = LCA(X,Y) , 然後 dir[x] + dir[y] - 2 * dir[lca], dir[u]表示點u到樹根的距離 */ #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> using namespace std; const int maxn=40010; struct note { int u,v,w,lca,next; }edge[maxn*2],edge1[805]; int head[maxn],ip,head1[maxn],ip1; int m,n; int father[maxn],vis[maxn],ance[maxn],dir[maxn]; void init() { memset(vis,0,sizeof(vis)); memset(dir,0,sizeof(dir)); memset(head,-1,sizeof(head)); memset(head1,-1,sizeof(head1)); ip=ip1=0; } void addedge(int u,int v,int w) { edge[ip].v=v,edge[ip].w=w,edge[ip].next=head[u],head[u]=ip++; } void addedge1(int u,int v) { edge1[ip1].u=u,edge1[ip1].v=v,edge1[ip1].lca=-1,edge1[ip1].next=head1[u],head1[u]=ip1++; } int Find(int x) { if(father[x]==x) return x; return father[x]=Find(father[x]); } void Union(int x,int y) { x=Find(x); y=Find(y); if(x!=y) father[y]=x; } void tarjan(int u) { vis[u]=1; ance[u]=father[u]=u; for(int i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].v; int w=edge[i].w; if(!vis[v]) { dir[v]=dir[u]+w; tarjan(v); Union(u,v); } } for(int i=head1[u];i!=-1;i=edge1[i].next) { int v=edge1[i].v; if(vis[v]) { edge1[i].lca=edge1[i^1].lca=ance[Find(v)]; } } } int main() { int T; scanf("%d",&T); while(T--) { init(); scanf("%d%d",&n,&m); for(int i=1;i<n;i++) { int u,v,w; scanf("%d%d%d",&u,&v,&w); addedge(u,v,w); addedge(v,u,w); } for(int i=0;i<m;i++) { int u,v; scanf("%d%d",&u,&v); addedge1(u,v); addedge1(v,u); } dir[1]=0; tarjan(1); for(int i=0;i<m;i++) { int s=i*2,u=edge1[s].u,v=edge1[s].v,lca=edge1[s].lca; printf("%d\n",dir[u]+dir[v]-2*dir[lca]); } } return 0; }
/** hdu 2586 LCA (線上演算法) 解題思路:轉為RMQ線上演算法求解,也是模板題 */ #include <stdio.h> #include <string.h> #include <algorithm> #include <iostream> #include <math.h> using namespace std; const int maxn=40010; const int maxm=25; int _pow[maxm],m,n; int head[maxn],ip; int ver[maxn*2],R[maxn*2],first[maxn],dir[maxn],dp[maxn*2][maxm],tot; bool vis[maxn]; void init() { memset(vis,false,sizeof(vis)); memset(head,-1,sizeof(head)); ip=0; } struct note { int v,w,next; }edge[maxn*2]; void addedge(int u,int v,int w) { edge[ip].v=v,edge[ip].w=w,edge[ip].next=head[u],head[u]=ip++; } void dfs(int u,int dep) { vis[u]=true; ver[++tot]=u,first[u]=tot,R[tot]=dep; for(int i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].v; int w=edge[i].w; if(!vis[v]) { dir[v]=dir[u]+w; dfs(v,dep+1); ver[++tot]=u,R[tot]=dep; } } } void ST(int len) { int k=(int)log((double)len)/(log(2.0)); for(int i=1;i<=len;i++) { dp[i][0]=i; } for(int j=1;j<=k;j++) { for(int i=1;i+_pow[j]-1<=len;i++) { int a=dp[i][j-1],b=dp[i+_pow[j-1]][j-1]; if(R[a]<R[b]) dp[i][j]=a; else dp[i][j]=b; } } } int RMQ(int x,int y) { int k=(int)log((double)(y-x+1)/log(2.0)); int a=dp[x][k],b=dp[y-_pow[k]+1][k]; if(R[a]<R[b]) return a; else return b; } int LCA(int u,int v) { int x=first[u],y=first[v]; if(x>y)swap(x,y); int res=RMQ(x,y); return ver[res]; } int main() { for(int i=0;i<maxn;i++)_pow[i]=(1<<i); int T; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); init(); for(int i=1;i<n;i++) { int u,v,w; scanf("%d%d%d",&u,&v,&w); addedge(u,v,w); addedge(v,u,w); } tot=0,dir[1]=0; dfs(1,1); ST(2*n-1); while(m--) { int u,v; scanf("%d%d",&u,&v); int lca=LCA(u,v); printf("%d\n",dir[u]+dir[v]-2*dir[lca]); } } return 0; }