#include <bits/stdc++.h>
#define long long long
using namespace std;
struct union_find {
vector<int> s, p;
union_find(int n = 0) : s(n, 1), p(n) {
iota(p.begin(), p.end(), 0);
}
int find(int u) {
if (p[u] != u) {
p[u] = find(p[u]);
}
return p[u];
}
bool merge(int u, int v) {
u = find(u);
v = find(v);
if (u == v) {
return false;
}
if (s[u] < s[v]) {
swap(u, v);
}
p[v] = u;
s[u] = s[u] + s[v];
return true;
}
bool same(int u, int v) {
return find(u) == find(v);
}
int size(int u) {
u = find(u);
return s[u];
}
};
/*
思维不难，但是一定一定要思路清晰。
1. 看到 k <= 20，想一想 bitmasks，如果想到了，不难想出一种简单的暴力：
我们枚举每一条需要选的边，然后强制把它加到树上（当然，如果出现环直接 pass）
再跑最小生成树，用边去限定我们那些没有确定边权的边（用了一个经典的结论），
时间复杂度为 2^k * mlogn，可以拿到约 56 分。
2. 我们发现，这整个过程中，总有一些边会加入到最小生成树，这也很好理解。
我们一共就 k 条可以选择的边，这些边能连起来的连通块就 k + 1 个，而图
一共有 n 个连通块，很显然一定是会有很多很多边需要连起来的。所以，我们
直接将所有 k 条边加入，然后再跑一边最小生成树，这个时候选择的边是一定
会一直选下去的。
3. 很自然的，我们缩一下点（这个使用并查集可以维护），我们现在就剩下了 k + 1
个点，再把剩余的那些原边跑一边最小生成树，这样子，我们就确保了多余的边只有
k 条，再算上原来的 k 条边，一共是 k * 2 条，瞎搞都可以。
*/
const int N = (int) 1E5;
const int M = (int) 3E5;
const int K = 20;
const int A = numeric_limits<int>::max();
int n, m, k;
bool t1[M];
array<int, 3> e1[M];
array<int, 2> e2[K];
array<int, 3> e3[K];
vector<int> e4[K + 1];
int a[N];
long e[K + 1];
void Kruskal1() {
union_find s(n);
for (int i = 0; i < k; i++) {
int u = e2[i][0], v = e2[i][1];
s.merge(u, v);
}
for (int i = 0; i < m; i++) {
int u = e1[i][1], v = e1[i][2];
t1[i] = s.merge(u, v);
}
}
int root;
void Kruskal2() {
union_find s1(n);
for (int i = 0; i < m; i++) {
int u = e1[i][1], v = e1[i][2];
if (t1[i] == true) {
s1.merge(u, v);
}
}
int c = 0;
vector<int> p(n, -1);
for (int i = 0; i < n; i++) {
if (s1.find(i) == i) {
p[i] = c;
c = c + 1;
}
}
assert(c == k + 1);
for (int i = 0; i < k; i++) {
int u = e2[i][0], v = e2[i][1];
u = p[s1.find(u)];
v = p[s1.find(v)];
e2[i] = {u, v};
}
int t = 0;
union_find s2(k + 1);
for (int i = 0; i < m; i++) {
int u = e1[i][1], v = e1[i][2], w = e1[i][0];
if (t1[i] == false) {
int x = p[s1.find(u)];
int y = p[s1.find(v)];
if (s2.merge(x, y)) {
e3[t] = {w, x, y};
t = t + 1;
}
}
}
assert(t == k);
for (int i = 0; i < n; i++) {
int u = p[s1.find(i)];
e[u] = e[u] + a[i];
}
root = p[s1.find(0)];
}
int p[K + 1], d[K + 1];
int g[K + 1];
long t[K + 1];
void DFS(int u) {
t[u] = e[u];
for (auto v : e4[u]) {
if (v != p[u]) {
p[v] = u;
d[v] = d[u] + 1;
DFS(v);
t[u] = t[u] + t[v];
}
}
}
void check(int u, int v, int w) {
while (u != v) {
if (d[u] > d[v]) {
swap(u, v);
}
g[v] = min(g[v], w);
v = p[v];
}
}
void solve() {
cin >> n >> m >> k;
for (int i = 0; i < m; i++) {
int u, v, w;
cin >> u >> v >> w;
u = u - 1, v = v - 1;
e1[i] = {w, u, v};
}
for (int i = 0; i < k; i++) {
int u, v;
cin >> u >> v;
u = u - 1, v = v - 1;
e2[i] = {u, v};
}
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(e1, e1 + m);
Kruskal1();
Kruskal2();
// for (int i = 0; i < k; i++) {
// cout << e2[i][0] << " " << e2[i][1] << "\n";
// }
// for (int i = 0; i < k; i++) {
// cout << e3[i][1] << " " << e3[i][2] << " " << e3[i][0] << "\n";
// }
// return;
long c = 0;
for (int S = 0; S < (1 << k); S++) {
union_find s(k + 1);
bool f = true;
for (int i = 0; i < k; i++) {
if (S >> i & 1) {
int u = e2[i][0], v = e2[i][1];
if (not s.merge(u, v)) {
f = false;
break;
}
}
}
if (not f) {
continue;
}
vector<bool> t3(k);
for (int i = 0; i < k; i++) {
int u = e3[i][1], v = e3[i][2];
t3[i] = s.merge(u, v);
}
for (int i = 0; i < k + 1; i++) {
e4[i] = vector<int>();
}
for (int i = 0; i < k; i++) {
if (S >> i & 1) {
int u = e2[i][0], v = e2[i][1];
e4[u].push_back(v);
e4[v].push_back(u);
}
}
for (int i = 0; i < k; i++) {
if (t3[i] == true) {
int u = e3[i][1], v = e3[i][2];
e4[u].push_back(v);
e4[v].push_back(u);
}
}
p[root] = -1;
d[root] = 0;
DFS(root);
// g[i] 表示节点 i 到节点 i 的父节点的边权最大是多少，初始的，为 MAX 即可
fill(g, g + k + 1, A);
for (int i = 0; i < k; i++) {
if (t3[i] == false) {
int u = e3[i][1], v = e3[i][2], w = e3[i][0];
check(u, v, w);
}
}
long b = 0;
for (int i = 0; i < k; i++) {
if (S >> i & 1) {
int u = e2[i][0], v = e2[i][1];
int x = d[u] > d[v] ? u : v;
b = b + t[x] * g[x];
}
}
c = max(c, b);
}
cout << c << "\n";
}
int32_t main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int t;
t = 1;
for (int i = 0; i < t; i++) {
solve();
}
return 0;
}
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Correct |
0 ms |
2396 KB |
Output is correct |
| 2 |
Correct |
1 ms |
2396 KB |
Output is correct |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Correct |
0 ms |
2396 KB |
Output is correct |
| 2 |
Correct |
1 ms |
2396 KB |
Output is correct |
| 3 |
Correct |
2 ms |
2396 KB |
Output is correct |
| 4 |
Runtime error |
2 ms |
4700 KB |
Execution killed with signal 6 |
| 5 |
Halted |
0 ms |
0 KB |
- |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Correct |
0 ms |
2396 KB |
Output is correct |
| 2 |
Correct |
1 ms |
2396 KB |
Output is correct |
| 3 |
Correct |
2 ms |
2396 KB |
Output is correct |
| 4 |
Runtime error |
2 ms |
4700 KB |
Execution killed with signal 6 |
| 5 |
Halted |
0 ms |
0 KB |
- |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Correct |
0 ms |
2396 KB |
Output is correct |
| 2 |
Correct |
1 ms |
2396 KB |
Output is correct |
| 3 |
Correct |
2 ms |
2396 KB |
Output is correct |
| 4 |
Runtime error |
2 ms |
4700 KB |
Execution killed with signal 6 |
| 5 |
Halted |
0 ms |
0 KB |
- |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Correct |
0 ms |
2396 KB |
Output is correct |
| 2 |
Correct |
1 ms |
2396 KB |
Output is correct |
| 3 |
Correct |
2 ms |
2396 KB |
Output is correct |
| 4 |
Runtime error |
2 ms |
4700 KB |
Execution killed with signal 6 |
| 5 |
Halted |
0 ms |
0 KB |
- |
