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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#483081 | #7103. Red Black Tree | lwenj | WA | 0ms | 15312kb | C++14 | 2.9kb | 2024-07-18 10:46:49 | 2024-07-18 10:46:49 |
Judging History
answer
#include <bits/stdc++.h>
#define MAXN ((int) 1e5)
#define MAXP 20
using namespace std;
int n, m, q;
bool flag[MAXN + 10];
vector<int> e[MAXN + 10], v[MAXN + 10];
// dis[i]:从根到节点 i 的距离
// cost[i]:不加入新红点时,节点 i 的代价
long long dis[MAXN + 10], cost[MAXN + 10];
// 在 dfs 序列上维护 rmq 求 lca
int clk, bgn[MAXN + 10];
int lg[MAXN * 2 + 10], f[MAXP][MAXN * 2 + 10];
void dfs(int sn, int fa) {
f[0][++clk] = sn; bgn[sn] = clk;
for (int i = 0; i < e[sn].size(); i++) {
int fn = e[sn][i];
if (fn == fa) continue;
dis[fn] = dis[sn] + v[sn][i];
if (flag[fn]) cost[fn] = 0;
else cost[fn] = cost[sn] + v[sn][i];
dfs(fn, sn);
f[0][++clk] = sn;
}
}
// rmq 预处理
void preLca() {
for (int p = 1; p < MAXP; p++)
for (int i = 1; i + (1 << p) - 1 <= clk; i++) {
int j = i + (1 << (p - 1));
if (dis[f[p - 1][i]] < dis[f[p - 1][j]])
f[p][i] = f[p - 1][i];
else
f[p][i] = f[p - 1][j];
}
}
// 求节点 x 和 y 的 lca
int lca(int x, int y) {
if (bgn[x] > bgn[y])
swap(x, y);
int p = lg[bgn[y] - bgn[x] + 1];
int a = f[p][bgn[x]], b = f[p][bgn[y] - (1 << p) + 1];
if (dis[a] < dis[b])
return a;
else
return b;
}
void solve() {
scanf("%d%d%d", &n, &m, &q);
memset(flag, 0, sizeof(bool) * (n + 3));
for (int i = 1; i <= m; i++) {
int x; scanf("%d", &x);
flag[x] = true;
}
for (int i = 1; i <= n; i++)
e[i].clear(), v[i].clear();
for (int i = 1; i < n; i++) {
int x, y, z; scanf("%d%d%d", &x, &y, &z);
e[x].push_back(y); v[x].push_back(z);
e[y].push_back(x); v[y].push_back(z);
}
clk = 0;
dfs(1, 0);
preLca();
while (q--) {
vector<int> vec;
int t; scanf("%d", &t);
while (t--) {
int x; scanf("%d", &x);
vec.push_back(x);
}
vec.push_back(0);
sort(vec.begin(), vec.end(), [&](int a, int b) {
return cost[a] > cost[b];
});
int anc = vec[0];
long long mx = dis[vec[0]], ans = cost[vec[0]];
// 枚举新的红点要覆盖几个节点
for (int i = 1; i + 1 < vec.size(); i++) {
anc = lca(anc, vec[i]);
mx = max(mx, dis[vec[i]]);
ans = min(ans, max(mx - dis[anc], cost[vec[i + 1]]));
}
printf("%lld\n", ans);
}
}
int main() {
lg[1] = 0;
for (int i = 2; i <= MAXN * 2; i++) lg[i] = lg[i >> 1] + 1;
int tcase; scanf("%d", &tcase);
while (tcase--) solve();
return 0;
}
/*
2
12 2 4
1 9
1 2 1
2 3 4
3 4 3
3 5 2
2 6 2
6 7 1
6 8 2
2 9 5
9 10 2
9 11 3
1 12 10
3 3 7 8
4 4 5 7 8
4 7 8 10 11
3 4 5 12
3 2 3
1 2
1 2 1
1 3 1
1 1
2 1 2
3 1 2 3
*/
详细
Test #1:
score: 0
Wrong Answer
time: 0ms
memory: 15312kb
input:
2 12 2 4 1 9 1 2 1 2 3 4 3 4 3 3 5 2 2 6 2 6 7 1 6 8 2 2 9 5 9 10 2 9 11 3 1 12 10 3 3 7 8 4 4 5 7 8 4 7 8 10 11 3 4 5 12 3 2 3 1 2 1 2 1 1 3 1 1 1 2 1 2 3 1 2 3
output:
4 5 3 10 0 0 1
result:
wrong answer 4th lines differ - expected: '8', found: '10'