P1525 [NOIP2010 提高组] 关押罪犯 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
种类并查集
#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
using namespace std;
using i64 = long long;
typedef pair<i64, i64> PII;
struct UFS {
int sz;
vector<int> rank, p;
void link(int x, int y) {
if (x == y)
return;
if (rank[x] > rank[y])
p[y] = x;
else
p[x] = y;
if (rank[x] == rank[y])
rank[y]++;
}
void init(int n) {
sz = n;
rank.resize(n + 1);
p.resize(n + 1);
for (int i = 0; i <= sz; i++) {
p[i] = i;
rank[i] = 0;
}
}
int find(int x) {
return x == p[x] ? x : (p[x] = find(p[x]));
}
void unin(int x, int y) {
link(find(x), find(y));
}
void compress() {
for (int i = 0; i <= sz; i++)
find(i);
}
};
//种类并查集 merge(y + n, x),merge(x + n, y)
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, m;
cin >> n >> m;
vector<pair<i64, PII>> g;
for (int i = 0; i < m; i ++) {
i64 u, v, w;
cin >> u >> v >> w;
g.emplace_back(w, PII(u, v));
}
sort(g.begin(), g.end(), greater<>());
UFS ufs;
ufs.init(2 * n);
i64 ans = 0;
for (int i = 0; i < m; i ++) {
auto [w, uv] = g[i];
auto [u, v] = uv;
if (ufs.find(u) == ufs.find(v) || ufs.find(u + n) == ufs.find(v + n)) {
ans = w;
break;
}
ufs.unin(u + n, v);
ufs.unin(v + n, u);
}
cout << ans << '\n';
return 0;
}
二分+二分图染色
二分最大的影响力,大于\(mid\)的罪犯就必须分开,然后看判定所有的罪犯是否可以分成两个集合
#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
using namespace std;
using i64 = long long;
typedef pair<i64, i64> PII;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, m;
cin >> n >> m;
vector g(n + 1, vector<PII>());
for (int i = 0; i < m; i ++) {
i64 u, v, w;
cin >> u >> v >> w;
g[u].emplace_back(v, w);
g[v].emplace_back(u, w);
}
auto check = [&](i64 mid) {
vector<int> col(n + 1, -1);
queue<int> Q;
for (int i = 1; i <= n; i ++) {
if (col[i] == -1) {
Q.push(i);
col[i] = 0;
while (Q.size()) {
auto u = Q.front();
Q.pop();
for (auto [v, w] : g[u]) {
if (w > mid) {
if (col[v] == -1) {
col[v] = col[u] ^ 1;
Q.push(v);
} else if (col[v] == col[u])
return false;
}
}
}
}
}
return true;
};
i64 l = 0, r = 1e9, ans = 0;
while (l <= r) {
i64 mid = (l + r) >> 1;
if (check(mid)) r = mid - 1, ans = mid;
else l = mid + 1;
}
cout << ans << '\n';
return 0;
}