477 字
2 分钟
P4281
题目
题解
弱化题目条件
如果是两个人的话,集合点肯定在这两个点的lca上
而这里是三个人,怎么办?
猜测发现一定在其中两个人的lca上
分别求出两两的lca,3种情况分别比较大小即可
代码
#include<bits/stdc++.h>const int N = 5e5 + 10;using namespace std;int n, m;int read(){ int f = 1, x = 0; 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 f * x;}
int head[N], nxt[N << 1], to[N << 1], cnt = 1;int fa[N][20];int depth[N];void add(int x, int y){ to[++cnt] = y; nxt[cnt] = head[x]; head[x] = cnt;}
void dfs(int u, int fath){ fa[u][0] = fath; depth[u] = depth[fath] + 1; for(int i = head[u]; i; i = nxt[i]) { int v = to[i]; if(v == fath) continue; dfs(v, u); }}
void initlca(){ for(int j = 1;j < 20;j++) { for(int i = 1;i <= n;i++) { fa[i][j] = fa[fa[i][j - 1]][j - 1]; } }}
int querylca(int x, int y){ if(depth[x] < depth[y]) swap(x, y);
for(int j = 19;j >= 0;j--) { if(depth[fa[x][j]] >= depth[y]) { x = fa[x][j]; } }
if(x == y) return x;
for(int j = 19;j >= 0;j--) { if(fa[x][j] != fa[y][j]) { x = fa[x][j], y = fa[y][j]; } }
return fa[x][0];}
int main(){ n = read(), m = read(); int a, b; for(int i = 1;i <= n - 1;i++) { a = read(), b = read(); add(a, b), add(b, a); } dfs(1, 0); initlca(); int x, y, z; while(m --) { x = read(), y = read(), z = read(); int xy = querylca(x, y); int yz = querylca(y, z); int xz = querylca(x, z); int xyz = querylca(xy, z); int yzx = querylca(yz, x); int xzy = querylca(xz, y); int xy_long = (-2) * depth[xy] + depth[x] + depth[y] - 2 * depth[xyz] + depth[z] + depth[xy]; int yz_long = (-2) * depth[yz] + depth[y] + depth[z] - 2 * depth[yzx] + depth[x] + depth[yz]; int xz_long = (-2) * depth[xz] + depth[x] + depth[z] - 2 * depth[xzy] + depth[y] + depth[xz];
if(xy_long <= yz_long && xy_long <= xz_long) { printf("%d %d\n", xy, xy_long); } else if(yz_long <= xy_long && yz_long <= xz_long) { printf("%d %d\n", yz, yz_long); } else if(xz_long <= xy_long && xz_long <= yz_long) { printf("%d %d\n", xz, xz_long); }
} return 0;}