BZOJ1602: [Usaco2008 Oct]牧场行走
2017-05-22 19:31
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题目传送门
题解:一个简单LCA即可 记录一下根到每个节点的距离
然后对于每个询问(a,b)的答案就是根到a的距离+根到b的距离-2*根到LCA(a,b)的距离
题解:一个简单LCA即可 记录一下根到每个节点的距离
然后对于每个询问(a,b)的答案就是根到a的距离+根到b的距离-2*根到LCA(a,b)的距离
#include<bits/stdc++.h> #define ll long long const int INF = 0x7fffffff; const double eps = 1e-5; const int maxn = 1e3 + 5; using namespace std; 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,q; int cnt,to[maxn<<1],nxt[maxn<<1],c[maxn<<1],head[maxn]; int deep[maxn],fa[maxn][18],dis[maxn],vis[maxn]; void add(int u,int v,int w) { to[++cnt] = v ; nxt[cnt] = head[u] ; head[u] = cnt; c[cnt] = w; to[++cnt] = u ; nxt[cnt] = head[v] ; head[v] = cnt; c[cnt] = w; } void dfs(int x,int f) { vis[x]=1; for(int i = 1; i <= 9; i++) { if(deep[x]<(1<<i)) break; fa[x][i] = fa[fa[x][i-1]][i-1]; } for(int i = head[x]; i; i = nxt[i]) { if(to[i] != f) { deep[to[i]] = deep[x]+1; dis[to[i]] = dis[x] + c[i]; fa[to[i]][0]=x; dfs(to[i],x); } } } int LCA(int a,int b) { if(deep[b] > deep[a]) swap(a,b); int t = deep[a] - deep[b]; for(int i = 0 ; i <= 9 ; i++) if(t&(1<<i)) a = fa[a][i]; for(int i = 0 ; i <= 9 ; i++) { while(fa[a][i]!=fa[b][i]) a=fa[a][i],b=fa[b][i]; } if(a==b) return a; else return fa[a][0]; } int main() { n = read() , q = read(); for(int i = 1 ; i < n ; i++) { int a = read() , b = read() , c = read(); add(a,b,c); } dfs(1,0); for(int i = 1 ; i <= q ; i++) { int a = read(), b =read(); printf("%d\n",dis[a]+dis[b]-(dis[LCA(a,b)]<<1)); } return 0; }
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