題目鏈接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5807
題目大意:給定一棵n個節點的樹,節點1爲根節點,樹上有m個節點是紅色的,剩下的節點都是黑色的。定義樹上每個節點的花費爲當前節點到達最近的爲紅色的祖先的距離。接下來有q次查詢,每次查詢給出k個節點,你允許將樹上對的任意一個節點染成紅色(僅對這次查詢有效),現在要求出如何染色,才能使得這k個點中花費是最大的節點的花費最小化,輸出這個最大花費。
題目思路:如果每次查詢中沒有染色的操作的話,我們可以通過dfs預處理出每個節點離它最近的爲紅色的祖先是哪個,距離是多少。接下來考慮加入染色操作的情況,由於是要使得最大花費最小化,所以我們就肯定是考慮對於一開始花費最大的點的祖先進行染色,這樣就可以降低最大值。根據這個思想,我們就可以先將這k個節點按照花費大小由大到小進行排序,在這之後枚舉對前p個點的lca進行染色的情況,求出最小的最大值即可,但要記得考慮邊界情況。
具體實現看代碼:
#include <bits/stdc++.h>
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pb push_back
#define MP make_pair
#define lowbit(x) x&-x
#define clr(a) memset(a,0,sizeof(a))
#define _INF(a) memset(a,0x3f,sizeof(a))
#define FIN freopen("in.txt","r",stdin)
#define IOS ios::sync_with_stdio(false)
#define fuck(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int MX = 1e5 + 7;
int n, m, q, _;
int red[MX];
struct edge {
int v, nxt;
ll w;
} E[MX << 1];
int head[MX], tot, sz;
int id[MX << 1], dep[MX << 1], first[MX], ST[MX << 1][20], vis[MX], vec[MX];
ll dis[MX], d[MX];
bool cmp(int a, int b) {
return d[a] > d[b];
}
void init(int _n) {
for (int i = 0; i <= _n; i++) {
head[i] = -1;
vis[i] = red[i] = dis[i] = d[i] = 0;
}
tot = sz = 0;
}
void add_edge(int u, int v, ll w) {
E[tot].v = v; E[tot].w = w; E[tot].nxt = head[u];
head[u] = tot++;
}
void dfs(int u, int deep, ll nw) {
if (red[u]) nw = 0;
d[u] = nw;
vis[u] = 1; id[++sz] = u;
first[u] = sz; dep[sz] = deep;
for (int i = head[u]; ~i; i = E[i].nxt) {
int v = E[i].v;
ll w = E[i].w;
if (vis[v]) continue;
dis[v] = dis[u] + w;
dfs(v, deep + 1, nw + w);
id[++sz] = u; dep[sz] = deep;
}
}
void ST_init(int _n) {
for (int i = 0; i <= _n; i++) ST[i][0] = i;
for (int j = 1; (1 << j) <= _n; j++) {
for (int i = 1; i + (1 << j) < _n; i++) {
int x = ST[i][j - 1], y = ST[i + (1 << (j - 1))][j - 1];
ST[i][j] = dep[x] < dep[y] ? x : y;
}
}
}
int RMQ(int l, int r) {
int k = 0;
while ((1 << (k + 1)) <= r - l + 1) k++;
int x = ST[l][k], y = ST[r - (1 << k) + 1][k];
return dep[x] < dep[y] ? x : y;
}
int LCA(int u, int v) {
int x = first[u], y = first[v];
if (x > y) swap(x, y);
return id[RMQ(x, y)];
}
int main() {
//FIN;
for (scanf("%d", &_); _; _--) {
scanf("%d%d%d", &n, &m, &q);
init(n);
for (int i = 1, x; i <= m; i++) {
scanf("%d", &x);
red[x] = 1;
}
for (int i = 1; i < n; i++) {
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
add_edge(u, v, w); add_edge(v, u, w);
}
dfs(1, 1, 0);
ST_init(sz);
while (q--) {
int k; scanf("%d", &k);
for (int i = 1; i <= k; i++) scanf("%d", &vec[i]);
sort(vec + 1, vec + k + 1, cmp);
ll ans = d[vec[1]], lastcnt = 0;
int lastlca = vec[1];
for (int i = 1; i <= k; i++) {
int lca = LCA(lastlca, vec[i]);
ll res1 = lastcnt + dis[lastlca] - dis[lca];
if (i > 1 && res1 >= d[vec[i - 1]]) break;
ll res2 = min(d[vec[i]], dis[vec[i]] - dis[lca]);
ll cnt = max(res1, res2);
if (cnt >= ans) break;
if (i + 1 <= k) ans = min(ans, max(cnt, d[vec[i + 1]]));
else ans = min(ans, cnt);
lastlca = lca; lastcnt = cnt;
}
printf("%lld\n", ans);
}
}
return 0;
}