点的距离
发布时间:2020-12-14 00:49:58 所属栏目:Linux 来源:网络整理
导读:#include iostream #include cstdio #include string using namespace std; int dfn[ 500001 ],f[ 500001 ][ 21 ]; int n,m,cnt,root; int head[ 500001 ],lg[ 500001 ]; struct tu { int v,nxt;} e[ 5000001 ]; void add( int u, int v) { e[ ++cnt].v = v;
#include <iostream> #include <cstdio> #include <string> using namespace std; int dfn[500001],f[500001][21]; int n,m,cnt,root; int head[500001],lg[500001]; struct tu { int v,nxt; } e[5000001]; void add(int u,int v) { e[++cnt].v = v; e[cnt].nxt = head[u]; head[u] = cnt; } void dfs(int now,int fa) { dfn[now] = dfn[fa] + 1; f[now][0] = fa; for (int i = 1; (1 << i) <= dfn[now]; i++) f[now][i] = f[f[now][i - 1]][i - 1]; for (int i = head[now]; i; i = e[i].nxt) { if (e[i].v != fa) dfs(e[i].v,now); } } int lca(int x,int y) { if (dfn[x] < dfn[y]) swap(x,y); while (dfn[x] > dfn[y]) { x = f[x][lg[dfn[x] - dfn[y]]]; } if (x == y) return x; for (int k = lg[dfn[x]]; k >= 0; k--) if (f[x][k] != f[y][k]) x = f[x][k],y = f[y][k]; return f[x][0]; } int main() { scanf("%d",&n); for (int i = 1; i <= n - 1; i++) { int x,y; scanf("%d%d",&x,&y); add(x,y); add(y,x); } dfs(1,0); lg[0] = -1; for (int i = 1; i <= n; i++) lg[i] = lg[i >> 1] + 1; //求log+1 // for(int i=1;i<=n;i++) // printf("%d ",lg[i]); scanf("%d",&m); for (int i = 1; i <= m; i++) { int x,&y); int mm = dfn[lca(x,y)]; printf("%dn",dfn[x] - mm + dfn[y] - mm); } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |