\(\verb|Luogu2860 [USACO06JAN]冗余路径Redundant Paths|\)
给定一个连通无向图,求至少加多少条边才能使得原图变为边双连通分量
\(1\leq n\leq5000,\ n-1\leq m\leq10^4\)
tarjan
边双无疑不用考虑,于是就可以边双缩点成一棵树
令现在要连的边为 \((u,\ v)\) ,点集 \(\{u->lca(u,\ v),\ v->lca(u,\ v)\}\) 将会变为一个新的点双,可以将他们看为一个新的点
可以贪心地连边使得每次连边后,不复存在的点尽量多,当只剩一个点时,原图就变成了一个双连通分量
如果 \(u\) 为非叶节点,显然不如将 \(u\) 子树中的一点 \(u‘\) 与 \(v\) 连接,于是 \(u,\ v\) 均为叶节点
若 \(lca(u,\ v)\) 为 \(root\) ,将会消去两个叶节点,否则只会消去一个叶节点,因此每次选择 \(lca(u,\ v)\) 为 \(root\) 的两个点,答案即为 \(叶节点的个数\lfloor\frac{\verb|叶节点的个数|+1}{2}\rfloor\)
时间复杂度 \(O(n+m)\)
代码
#include <bits/stdc++.h>
using namespace std;
#define nc getchar()
const int maxn = 5010;
int n, m, tot, h[maxn], bl[maxn], dfn[maxn], low[maxn], deg[maxn]; bool vis[maxn], cut[maxn << 1];
struct edges {
int nxt, to;
edges(int x = 0, int y = 0) : nxt(x), to(y) {}
} e[maxn << 1];
inline int read() {
int x = 0; char c = nc;
while (c < 48) c = nc;
while (c > 47) x = x * 10 + c - 48, c = nc;
return x;
}
void addline(int u, int v) {
static int cnt = 1;
e[++cnt] = edges(h[u], v), h[u] = cnt;
}
void tarjan(int u, int f) {
static int now;
dfn[u] = low[u] = ++now;
for (int i = h[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (!dfn[v]) {
tarjan(v, u);
low[u] = min(low[u], low[v]);
if (dfn[u] < low[v]) cut[i] = cut[i ^ 1] = 1;
} else if (v != f) {
low[u] = min(low[u], dfn[v]);
}
}
}
void dfs(int u) {
vis[u] = 1, bl[u] = tot;
for (int i = h[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (!cut[i] && !vis[v]) dfs(v);
}
}
int main() {
n = read(), m = read();
for (int i = 1; i <= m; i++) {
int u = read(), v = read();
addline(u, v), addline(v, u);
}
tarjan(1, 0);
for (int i = 1; i <= n; i++) {
if (!vis[i]) tot++, dfs(i);
}
for (int u = 1; u <= n; u++) {
for (int i = h[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (bl[u] != bl[v]) deg[bl[v]]++;
}
}
int ans = 0;
for (int i = 1; i <= tot; i++) {
ans += deg[i] == 1;
}
printf("%d", (ans + 1) >> 1);
return 0;
}
原文地址:https://www.cnblogs.com/Juanzhang/p/10375244.html
时间: 2024-10-09 21:42:26