题目链接
给定一棵n个点的树,边具有边权。要求作以下操作:
- DIST a b 询问点a至点b路径上的边权之和
- KTH a b k 询问点a至点b有向路径上的第k个点的编号
有多组测试数据,每组数据以DONE结尾。
其实,一个LCA也可以维护上述两个东西,求路径和可以用差分的思想来做,求kth可以通过判断在lca的左边还是右边来求。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define eps 1e-4
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e4 + 7;
int N, head[maxN], cnt;
struct Eddge
{
int nex, to, val;
Eddge(int a=-1, int b=0, int c=0):nex(a), to(b), val(c) {}
} edge[maxN << 1];
inline void addEddge(int u, int v, int w)
{
edge[cnt] = Eddge(head[u], v, w);
head[u] = cnt++;
}
inline void _add(int u, int v, int w) { addEddge(u, v, w); addEddge(v, u, w); }
int siz[maxN], Wson[maxN], root[maxN], nw[maxN], deep[maxN], father[maxN][15], LOG_2[maxN];
void dfs_1(int u, int fa)
{
Wson[u] = 0; siz[u] = 1; root[u] = fa; deep[u] = deep[fa] + 1; father[u][0] = fa;
for(int i=0; (1 << (i + 1)) < N; i++) father[u][i + 1] = father[father[u][i]][i];
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(v == fa) continue;
nw[v] = edge[i].val;
dfs_1(v, u);
if(siz[v] > siz[Wson[u]]) Wson[u] = v;
}
}
int top[maxN], dfn[maxN], tot, W[maxN];
void dfs_2(int u, int topy)
{
top[u] = topy;
dfn[u] = ++tot;
W[tot] = nw[u];
if(Wson[u]) dfs_2(Wson[u], topy);
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(v == root[u] || v == Wson[u]) continue;
dfs_2(v, v);
}
}
int tree[maxN << 2];
inline void pushup(int rt) { tree[rt] = tree[lsn] + tree[rsn]; }
void buildTree(int rt, int l, int r)
{
if(l == r) { tree[rt] = W[l]; return; }
int mid = HalF;
buildTree(Lson); buildTree(Rson);
pushup(rt);
}
int query(int rt, int l, int r, int ql, int qr)
{
if(ql <= l && qr >= r) return tree[rt];
int mid = HalF;
if(qr <= mid) return query(QL);
else if(ql > mid) return query(QR);
else return query(QL) + query(QR);
}
inline int Range_Query(int u, int v)
{
int ans = 0;
while(top[u] ^ top[v])
{
if(deep[top[u]] < deep[top[v]]) swap(u, v);
ans += query(1, 1, N, dfn[top[u]], dfn[u]);
u = root[top[u]];
}
if(deep[u] > deep[v]) swap(u, v);
if(dfn[u] < dfn[v]) ans += query(1, 1, N, dfn[u] + 1, dfn[v]);
return ans;
}
inline int kth(int u, int v, int k)
{
int x = u, y = v;
if(deep[x] < deep[y]) swap(x, y);
int det = deep[x] - deep[y];
for(int i=LOG_2[det]; i>=0; i--)
{
if((det >> i) & 1) x = father[x][i];
}
int lca;
if(x == y) lca = x;
else
{
for(int i=LOG_2[N]; i>=0; i--)
{
if(father[x][i] ^ father[y][i])
{
x = father[x][i];
y = father[y][i];
}
}
lca = father[x][0];
}
int len = deep[u] - deep[lca] + 1;
if(k <= len)
{
k--;
for(int i=LOG_2[k]; i>=0; i--)
{
if((k >> i) & 1) u = father[u][i];
}
}
else
{
k = deep[u] + deep[v] - deep[lca] * 2 + 1 - k;
u = v;
for(int i=LOG_2[k]; i>=0; i--)
{
if((k >> i) & 1) u = father[u][i];
}
}
return u;
}
inline void init()
{
cnt = tot = 0;
for(int i=1; i<=N; i++) head[i] = -1;
}
int main()
{
for(int i = 1, j = 2, k = 0; i<maxN; i++)
{
if(i == j) { j <<= 1; k++; }
LOG_2[i] = k;
}
int T; scanf("%d", &T);
while(T--)
{
scanf("%d", &N);
init();
for(int i=1, u, v, w; i<N; i++)
{
scanf("%d%d%d", &u, &v, &w);
_add(u, v, w);
}
dfs_1(1, 0);
dfs_2(1, 1);
buildTree(1, 1, N);
char op[10];
int a, b, k;
while(scanf("%s", op) && op[1] ^ 'O')
{
if(op[0] == 'D')
{
scanf("%d%d", &a, &b);
printf("%d\n", Range_Query(a, b));
}
else
{
scanf("%d%d%d", &a, &b, &k);
printf("%d\n", kth(a, b, k));
}
}
}
return 0;
}