LCA
题意：给一个无根树，有q个询问，每个询问3个点，问将这3个点连起来，距离最短是多少，LCA的模板题，分别求LCA(X,Y),LCA(X,Z),LCA(Y,Z)，和对应的距离，然后3个距离相加再除以2就是这个询问的结果
对于一对点,x,y, lca = LCA(x,y) ， 那么点x到点y的距离为 dir[x] + dir[y] - 2 * dir[lca] ; 其中dir[u] 表示点u到树根的距离
由于是模板题，只给代码，详细的讲解可以在学习笔记里面找《LCA与RMQ》
在线算法：LCA转RMQ
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <vector> using namespace std; const int N = 50010; const int M = 25; int _pow[M]; int n,tot; bool vis[N]; int ver[2*N],R[2*N],first[N],dir[N]; int dp[2*N][M]; //这个数组记得开到2*N，因为遍历后序列长度为2*n-1 struct node { int v,w; node(int a, int b) { v = a; w = b;} }; vector<node>g[N]; void dfs(int u ,int dep) { vis[u] = true; ver[++tot] = u; first[u] = tot; R[tot] = dep; for(int i=0; i<g[u].size(); i++) if( !vis[g[u][i].v] ) { int v = g[u][i].v , w = g[u][i].w; 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<M; i++) _pow[i] = (1<<i); int cas = 0; while(scanf("%d",&n)!=EOF) { if(cas++) printf("\n"); for(int i=0; i<n; i++) g[i].clear() , vis[i] = false; for(int i=1; i<n; i++) { int u,v,w; scanf("%d%d%d",&u,&v,&w); g[u].push_back(node(v,w)); g[v].push_back(node(u,w)); } tot = 0; dir[0] = 0; dfs(0,1); /* printf("节点 "); for(int i=1; i<=2*n-1; i++) printf("%d ",ver[i]); cout << endl; printf("深度 "); for(int i=1; i<=2*n-1; i++) printf("%d ",R[i]); cout << endl; printf("首位 "); for(int i=0; i<n; i++) printf("%d ",first[i]); cout << endl; printf("距离 "); for(int i=0; i<n; i++) printf("%d ",dir[i]); cout << endl; */ ST(2*n-1); int q; scanf("%d",&q); while(q--) { int x,y,z; scanf("%d%d%d",&x,&y,&z); int lca1 = LCA(x,y); int res1 = dir[x] + dir[y] - 2*dir[lca1]; int lca2 = LCA(x,z); int res2 = dir[x] + dir[z] - 2*dir[lca2]; int lca3 = LCA(y,z); int res3 = dir[y] + dir[z] - 2*dir[lca3]; printf("%d\n",(res1 + res2 + res3)/2); } } return 0; }
离线算法：Tarjan
#include <iostream> #include <cstdio> #include <cstring> #include <vector> using namespace std; const int N = 50010; const int M = 420010; int n,tot; int dir[N]; int fa[N]; //并查集 int ance[N]; //并查集的祖先 struct node { int v,w; node(int a, int b) { v=a; w=b; } }; vector<node>g[N]; bool vis[N]; int head[N]; struct ask { int u,v,lca,c,next; }ea[M]; int find(int x) {//并查集查找元素且路径压缩 return x == fa[x] ? x : fa[x] = find(fa[x]); } void unionset(int x ,int y) {//合并元素x和元素y所在的集合 fa[find(y)] = find(x); } void Tarjan(int u) { vis[u] = true; fa[u] = u; //以点u建立集合，u为代表元素 //ance[find(u)] = u; //该集合的祖先也是u自己 ance[u] = u; for(int i=0; i<g[u].size(); i++) if( !vis[g[u][i].v]) { int v = g[u][i].v , w = g[u][i].w; dir[v] = dir[u] + w; Tarjan(v); unionset(u,v); //将儿子所在的集合并到自己的集合里 //ance[find(u)] = u; //保证自己所在的那个集合的祖先还是自己 } for(int k=head[u]; k!=-1; k=ea[k].next) if( vis[ea[k].v] ) { int v = ea[k].v; ea[k^1].lca = ea[k].lca = ance[find(v)]; } } inline void add_ask(int u , int v ,int c) { ea[tot].u = u; ea[tot].v = v; ea[tot].c = c; ea[tot].lca = -1; ea[tot].next = head[u]; head[u] = tot++; u = u^v; v = u^v; u = u^v; ea[tot].u = u; ea[tot].v = v; ea[tot].c = c; ea[tot].lca = -1; ea[tot].next = head[u]; head[u] = tot++; } int main() { int cas = 0; while(scanf("%d",&n)!=EOF) { if(cas++) puts(""); for(int i=0; i<n; i++) g[i].clear() , vis[i] = false; for(int i=1; i<n; i++) { int u,v,w; scanf("%d%d%d",&u,&v,&w); g[u].push_back(node(v,w)); g[v].push_back(node(u,w)); } int q; tot = 0; memset(head,-1,sizeof(head)); scanf("%d",&q); for(int i=0; i<q; i++) { int x,y,z; //要处理的询问包括 LCA(x,y),LCA(x,z),LCA(y,z) scanf("%d%d%d",&x,&y,&z); add_ask(x,y,i); add_ask(x,z,i); add_ask(y,z,i); } dir[0] = 0; Tarjan(0); for(int i=0; i<q; i++) { int s = i*6; int x = ea[s].u , y = ea[s].v , z = ea[s+2].v; int lca1 = ea[s].lca; //LCA(x,y) int lca2 = ea[s+2].lca; //LCA(x,z) int lca3 = ea[s+4].lca; //LCA(y,z) int res1 = dir[x] + dir[y] - 2*dir[lca1]; int res2 = dir[x] + dir[z] - 2*dir[lca2]; int res3 = dir[y] + dir[z] - 2*dir[lca3]; printf("%d\n",(res1+res2+res3)/2); } } return 0; }