ACM-ICPC 2018 青岛赛区网络预赛 B. Red Black Tree (LCA、二分)
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传送门：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5807
题意：
给出一棵树，根节点为1。每条边有一个距离，树上有m个点为红色的点，其余为黑色，每个点的权值为到其最近红色祖先的距离。有q次询问，每次给出一个点集，问在树上涂红一个点后，点集中所有点的最大值的最小值是多少。
思路：
预处理每个点到根的距离cost、到最近红色祖先的距离cost_t和ST表。
对于每次询问，按cost_t从大到小排序，在0~cost_t[0]范围内二分答案，对所有大于答案的点求它们的公共祖先(利用ST表可以O(1)求两点的公共祖先)，将其涂红，之后计算每个大于答案的点的新权值是否小于答案。
AC代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<utility>
#include<algorithm>
#include<utility>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<cmath>
#include<map>
#include<ctime>
#include<functional>
#include<bitset>
#define P pair<int,int>
#define ll long long
#define ull unsigned long long
#define lson id*2,l,mid
#define rson id*2+1,mid+1,r
#define ls id*2
#define rs (id*2+1)
#define Mod(a,b) a<b?a:a%b+b
#define cl0(a) memset(a,0,sizeof(a))
#define cl1(a) memset(a,-1,sizeof(a))
using namespace std;
const ll M = 1e9 + 7;
const ll INF = 1e15;
const int N = 4010;
const double _e = 10e-6;
const int maxn = 1e5 + 10;
const int dx[4] = { 0,0,1,-1 }, dy[4] = { 1,-1,0,0 };
const int _dx[8] = { -1,-1,-1,0,0,1,1,1 }, _dy[8] = { -1,0,1,-1,1,-1,0,1 };
int x, y;
int t, n, m, q, cc;
bool red[maxn];
struct node
{
int nex, to;
ll c;
}e[maxn * 2];
int beg[2 * maxn], e_max;
int pos[2 * maxn], T[2 * maxn], tot, rmq[2 * maxn];
ll cost[maxn], cost_t[maxn];
struct ST
{
int mm[2 * maxn];
int dp[2 * maxn][20];
void init(int n)
{
mm[0] = -1;
for (int i = 1; i <= n; i++) {
mm[i] = ((i&(i - 1)) == 0) ? mm[i - 1] + 1 : mm[i - 1];
dp[i][0] = i;
}
for (int j = 1; j <= mm[n]; j++)
for (int i = 1; i + (1 << j) - 1 <= n; i++)
dp[i][j] = rmq[dp[i][j - 1]] < rmq[dp[i + (1 << (j - 1))][j - 1]] ? dp[i][j - 1] : dp[i + (1 << (j - 1))][j - 1];
}
int query(int a, int b)
{
if (a > b)swap(a, b);
int k = mm[b - a + 1];
return rmq[dp[a][k]] <= rmq[dp[b - (1 << k) + 1][k]] ? dp[a][k] : dp[b - (1 << k) + 1][k];
}
}st;
void init()
{
memset(beg, -1, sizeof beg);
e_max = 0; tot = 1;
}
void add_edge(int s, int t, int cost)
{
e[e_max].to = t; e[e_max].c = cost;
e[e_max].nex = beg[s]; beg[s] = e_max++;
}
void dfs(int u, int pre, int dis, ll sum, ll sum1)
{
if (red[u])sum1 = 0;
cost[u] = sum; cost_t[u] = sum1;
pos[u] = tot; T[tot] = u; rmq[tot++] = dis;
for (int i = beg[u]; ~i; i = e[i].nex) {
int v = e[i].to;
if (v == pre) continue;
dfs(v, u, dis + 1, sum + e[i].c, sum1 + e[i].c);
T[tot] = u;
rmq[tot++] = dis;
}
}
void lca_init()
{
dfs(1, -1, 1, 0, 0);
st.init(tot - 1);
}
int query(int s, int t)
{
return T[st.query(pos[s], pos[t])];
}
int a[maxn];
int nn;
bool cmp(int a, int b)
{
return cost_t[a] > cost_t[b];
}
bool check(ll val)
{
if (cost_t[a[0]] <= val)
return true;
int lca = a[0];
for (int i = 1; i < nn; i++) {
if (cost_t[a[i]] <= val)
break;
lca = query(lca, a[i]);
}
for (int i = 0; i < nn; i++) {
if (cost_t[a[i]] <= val)
return true;
if (cost[a[i]] - cost[lca] > val)
return false;
}
return true;
}
int main()
{
scanf("%d", &t);
while (t--) {
init();
memset(red, false, sizeof(red));
scanf("%d%d%d", &n, &m, &q);
while (m--) {
scanf("%d", &x);
red[x] = true;
}
for (int i = 1; i < n; i++) {
scanf("%d%d%d", &x, &y, &cc);
add_edge(x, y, cc); add_edge(y, x, cc);
}
lca_init();
while (q--) {
scanf("%d", &nn);
for (int i = 0; i < nn; i++)
scanf("%d", &a[i]);
sort(a, a + nn, cmp);
ll l = 0, r = cost_t[a[0]];
while (l < r) {
ll mid = (l + r) / 2;
if (check(mid))
r = mid;
else
l = mid + 1;
}
printf("%lld\n", l);
}
}
return 0;
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