标签：amp inf rect ssi ring mil HERE 距离 one
You are given a tree (an undirected acyclic connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. Each edge has an integer value assigned to it, representing its length.
We will ask you to perfrom some instructions of the following form:
Example:
N = 6
1 2 1 // edge connects node 1 and node 2 has cost 1
2 4 1
2 5 2
1 3 1
3 6 2
Path from node 4 to node 6 is 4 -> 2 -> 1 -> 3 -> 6
DIST 4 6 : answer is 5 (1 + 1 + 1 + 2 = 5)
KTH 4 6 4 : answer is 3 (the 4-th node on the path from node 4 to node 6 is 3)
The first line of input contains an integer t, the number of test cases (t <= 25). t test cases follow.
For each test case:
There is one blank line between successive tests.
For each "DIST" or "KTH" operation, write one integer representing its result.
Print one blank line after each test.
Input: 1 6 1 2 1 2 4 1 2 5 2 1 3 1 3 6 2 DIST 4 6 KTH 4 6 4 DONE Output: 5 3
看到树上两点距离很容易想到LCA，
对于第K个点我们同样可以倍增解决；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 9999973;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == ‘-‘) f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
struct node {
int u, v, w, nxt;
}e[maxn];
int head[maxn];
int tot;
int n;
int dis[maxn], dep[maxn];
int fa[maxn][20];
void init() {
ms(e); ms(head); tot = 0; ms(dis); ms(dep);
ms(fa);
}
void addedge(int u, int v, int w) {
e[++tot].u = u; e[tot].v = v; e[tot].nxt = head[u]; e[tot].w = w;
head[u] = tot;
}
void dfs(int rt) {
for (int i = 1; i <= (int)log(n) / log(2) + 1; i++)
fa[rt][i] = fa[fa[rt][i - 1]][i - 1];
for (int i = head[rt]; i; i = e[i].nxt) {
int v = e[i].v;
if (v == fa[rt][0])continue;
fa[v][0] = rt; dep[v] = dep[rt] + 1;
dis[v] = dis[rt] + e[i].w;
dfs(v);
}
}
int LCA(int x, int y) {
if (dep[x] > dep[y])swap(x, y);
for (int i = (int)log(n) / log(2) + 1; i >= 0; i--) {
if (dep[fa[y][i]] >= dep[x])y = fa[y][i];
}
if (x == y)return x;
for (int i = (int)log(n) / log(2) + 1; i >= 0; i--) {
if (fa[x][i] != fa[y][i]) {
x = fa[x][i]; y = fa[y][i];
}
}
return fa[x][0];
}
int main()
{
// ios::sync_with_stdio(0);
int T = rd();
while (T--) {
n = rd();
init();
for (int i = 1; i < n; i++) {
int u = rd(), v = rd(), w = rd();
addedge(u, v, w); addedge(v, u, w);
}
dfs(1);
char op[20];
while (rdstr(op) != EOF && op[1] != ‘O‘) {
if (op[1] == ‘I‘) {
int u = rd(), v = rd();
// cout << dis[u] << ‘ ‘ << dis[v] << ‘ ‘ << dis[LCA(u, v)] << endl;
printf("%d\n", dis[u] + dis[v] - 2 * dis[LCA(u, v)]);
}
else {
int u = rd(), v = rd(), k = rd();
int root = LCA(u, v);
int ans;
if (dep[u] - dep[root] + 1 >= k) {
ans = dep[u] - k + 1;
int i;
for (i = 0; (1 << i) <= dep[u]; i++); i--;
for (int j = i; j >= 0; j--) {
if (dep[u] - (1 << j) >= ans)u = fa[u][j];
}
printf("%d\n", u);
}
else {
ans = dep[root] + k - (dep[u] - dep[root] + 1);
int i;
for (i = 0; (1 << i) <= dep[v]; i++); i--;
for (int j = i; j >= 0; j--) {
if (dep[v] - (1 << j) >= ans)v = fa[v][j];
}
printf("%d\n", v);
}
}
}
}
return 0;
}
标签：amp inf rect ssi ring mil HERE 距离 one
原文地址：https://www.cnblogs.com/zxyqzy/p/10355532.html
