poj——1986 Distance Queries
Time Limit: 2000MS | Memory Limit: 30000K | |
Total Submissions: 14392 | Accepted: 5066 | |
Case Time Limit: 1000MS |
Description
Input
* Line 2+M: A single integer, K. 1 <= K <= 10,000
* Lines 3+M..2+M+K: Each line corresponds to a distance query and contains the indices of two farms.
Output
Sample Input
7 6 1 6 13 E 6 3 9 E 3 5 7 S 4 1 3 N 2 4 20 W 4 7 2 S 3 1 6 1 4 2 6
Sample Output
13 3 36
Hint
Source
#include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define N 51000 using namespace std; int n,m,x,y,z,t,tot,ans; int fa[N],dis[N],top[N],deep[N],size[N],head[N]; struct Edge { int from,to,dis,next; }edge[N<<]; int read() { ,f=; char ch=getchar(); ; ch=getchar();} +ch-'; ch=getchar();} return x*f; } int add(int x,int y,int z) { tot++; edge[tot].to=y; edge[tot].dis=z; edge[tot].next=head[x]; head[x]=tot; } int dfs(int x) { size[x]=; deep[x]=deep[fa[x]]+; for(int i=head[x];i;i=edge[i].next) { int to=edge[i].to; if(fa[x]==to) continue; dis[to]=dis[x]+edge[i].dis; fa[to]=x,dfs(to);size[x]+=size[to]; } } int dfs1(int x) { ; if(!top[x]) top[x]=x; for(int i=head[x];i;i=edge[i].next) { int to=edge[i].to; if(fa[x]!=to&&size[t]<size[to]) t=to; } if(t) top[t]=top[x],dfs1(t); for(int i=head[x];i;i=edge[i].next) { int to=edge[i].to; if(fa[x]!=to&&to!=t) dfs1(to); } } int lca(int x,int y) { for(;top[x]!=top[y];x=fa[top[x]]) if(deep[top[x]]<deep[top[y]]) swap(x,y); if(deep[x]>deep[y]) swap(x,y); return x; } int main() { n=read(),m=read(); ;i<=m;i++) { x=read(),y=read(),z=read(); add(x,y,z),add(y,x,z); } dfs(),dfs1(); t=read(); ;i<=t;i++) { x=read(),y=read(); ans=dis[x]+dis[y]-*dis[lca(x,y)]; printf("%d\n",ans); } ; }
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