【LG5022】[NOIP2018]旅行
题面
题解
首先考虑一棵树的部分分怎么打
直接从根节点开始(dfs),依次选择编号最小的儿子即可
而此题是一个基环树
怎么办呢?
可以断掉环上的一条边,这样就变为一棵树了
再用上面的方法做即可
(tips) (:)
断环上的边,其实可以直接用(tarjan)把桥求出来
不是桥的就是环上的边
考场上的代码有点乱
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
inline int gi() {
register int data = 0, w = 1;
register char ch = 0;
while (ch != '-' && !isdigit(ch)) ch = getchar();
if (ch == '-') w = -1, ch = getchar();
while (isdigit(ch)) data = data * 10 + ch - '0', ch = getchar();
return w * data;
}
#define MAX_N 5005
vector<int> G[MAX_N];
struct Graph { int to, next; } e[MAX_N << 1]; int fir[MAX_N], e_cnt;
void clearGraph(){ memset(fir, -1, sizeof(fir)); }
void Add_Edge(int u, int v) { e[e_cnt] = (Graph){v, fir[u]}; fir[u] = e_cnt++; }
struct Edge {
int to, id;
bool operator < (const Edge &rhs) const { return to < rhs.to; }
} ;
vector<Edge> rG[MAX_N];
int N, M;
namespace cpp1 {
void dfs(int x, int f) {
printf("%d ", x);
for (int i = 0, sz = rG[x].size(); i < sz; i++) {
int v = rG[x][i].to;
if (v == f) continue;
dfs(v, x);
}
}
void main() { dfs(1, 0); putchar('
'); }
}
namespace cpp2 {
int dfn[MAX_N], low[MAX_N], tim;
bool bridge[MAX_N << 1];
void tarjan(int x, int id) {
dfn[x] = low[x] = ++tim;
for (int i = fir[x]; ~i; i = e[i].next) {
int v = e[i].to;
if (!dfn[v]) {
tarjan(v, i), low[x] = min(low[x], low[v]);
if (low[v] > dfn[x]) bridge[i] = bridge[i ^ 1] = 1;
} else if (i != (id ^ 1)) low[x] = min(low[x], dfn[v]);
}
}
int ans[MAX_N], tmp[MAX_N], cnt;
bool used[MAX_N << 1];
void dfs(int x, int fa) {
tmp[++cnt] = x;
for (int i = 0, sz = rG[x].size(); i < sz; i++) {
int v = rG[x][i].to, id = rG[x][i].id;
if (v == fa || used[id]) continue;
dfs(v, x);
}
}
bool check() {
for (int i = 1; i <= N; i++) {
if (ans[i] > tmp[i]) return 1;
else if (ans[i] < tmp[i]) return 0;
}
return 0;
}
void main() {
tarjan(1, -1);
for (int i = 1; i <= N; i++) ans[i] = N + 1;
for (int i = 0; i < e_cnt; i += 2) {
if (bridge[i]) continue;
used[i] = used[i ^ 1] = 1; cnt = 0;
dfs(1, 0);
if (check()) for (int j = 1; j <= N; j++) ans[j] = tmp[j];
used[i] = used[i ^ 1] = 0;
}
for (int i = 1; i <= N; i++) printf("%d ", ans[i]);
putchar('
');
}
}
bool used[MAX_N][MAX_N];
int main () {
N = gi(), M = gi();
for (int i = 1; i <= M; i++) {
int u = gi(), v = gi();
G[u].push_back(v), G[v].push_back(u);
}
for (int i = 1; i <= N; i++) sort(G[i].begin(), G[i].end());
clearGraph();
for (int i = 1; i <= N; i++)
for (int j = G[i].size() - 1; j >= 0; j--) {
int v = G[i][j]; if (used[i][v]) continue;
Add_Edge(i, v), Add_Edge(v, i);
used[i][v] = used[v][i] = 1;
}
for (int x = 1; x <= N; x++) {
for (int i = fir[x]; ~i; i = e[i].next) {
int v = e[i].to;
rG[x].push_back((Edge){v, i});
}
}
for (int i = 1; i <= N; i++) sort(rG[i].begin(), rG[i].end());
if (N - 1 == M) cpp1::main();
else cpp2::main();
return 0;
}