Codeforces Round #731 (Div. 3) G. How Many Paths? 强连通分量缩点+拓扑
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原题链接:https://codeforces.ml/contest/1547/problem/G
题意
给你一个n个点,m条边的有向图,可能存在自环。定义了四种状态:
- 无法从1到v,输出-1
- 有且仅有一条路径到v,输出1
- 有不只一条路径到v,输出2
- 有无数条路径到v,输出-1
输出 [ 1 , n ] [1,n] [1,n]所有点的状态
分析
遇到有环的图,最常见的做法就是缩点然后在DAG图上跑拓扑。本题的状态转移也是相当明显的,基本上没有什么思维量,看到tag应该就可以写出来了,只需要讨论一下四种状态之间的转移就可以。
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int ul;
typedef pair<ll, ll> PII;
const ll inf = 1e18;
const int N = 4e5 + 10;
const int M = 1e6 + 10;
const ll mod = 1e9 + 7;
const double eps = 1e-8;
#define lowbit(i) (i & -i)
#define Debug(x) cout << (x) << endl
#define fi first
#define se second
#define mem memset
#define endl '\\n'
namespace StandardIO {
template<typename T>
inline void read(T &x) {
x = 0; T f = 1;
char c = getchar();
for (; c < '0' || c > '9'; c = getchar()) if (c == '-') f = -1;
for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
x *= f;
}
template<typename T>
inline void write(T x) {
if (x < 0) putchar('-'), x *= -1;
if (x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
}
int in[N], f[N], ff[N], cir[N];
int dfn[N], low[N], idx, tp, in_stk[N], vis[N], sd[N], scc;
vector<int> g[N], G[N], bel[N];
void tarjan(int x) {
low[x] = dfn[x] = ++idx;
vis[x] = 1;
in_stk[++tp] = x;
for (auto v : g[x]) {
if (!dfn[v]) {
tarjan(v);
low[x] = min(low[x], low[v]);
}
else if (vis[v]) {
low[x] = min(low[x], dfn[v]);
}
}
if (low[x] == dfn[x]) {
int y;
++scc;
while(y = in_stk[tp--]) {
sd[y] = scc;
bel[scc].push_back(y);
vis[y] = 0;
if (x == y) break;
}
}
}
void tupo() {
queue<int> que;
for (int i = 1; i <= scc; i++) {
if (!in[i]) que.push(i);
}
f[sd[1]] = 1;
while (que.size()) {
int now = que.front();
que.pop();
if (cir[now] && f[now] != 0) f[now] = -1;
for (auto v : G[now]) {
if (f[now] == 1) {
if (!f[v]) {
f[v] = 1;
} else if (f[v] == 1) {
f[v] = 2;
}
} else if (f[now] == 2) {
if (!f[v] || f[v] == 1) {
f[v] = 2;
}
} else if (f[now] == -1) {
f[v] = -1;
}
if (--in[v] == 0) {
que.push(v);
}
}
}
}
inline void solve() {
int T; cin >> T; while (T--) {
int n, m; cin >> n >> m;
for (int i = 1; i <= n; i++) {
g[i].clear(), G[i].clear(), bel[i].clear();
in[i] = f[i] = vis[i] = dfn[i] = low[i] = cir[i] = ff[i] = sd[i] = 0;
}
tp = idx = scc = 0;
for (int i = 1; i <= m; i++) {
int u, v; cin >> u >> v;
g[u].push_back(v);
}
for (int i = 1; i <= n; i++) if (!dfn[i]) tarjan(i);
for (int i = 1; i <= scc; i++) {
for (auto u : bel[i]) {
for (auto v : g[u]) {
if (sd[v] != sd[u]) {
G[sd[u]].push_back(sd[v]);
in[sd[v]]++;
} else {
cir[i] = 1;
}
}
}
}
tupo();
for (int i = 1; i <= scc; i++) {
for (auto it : bel[i]) ff[it] = f[i];
}
for (int i = 1; i <= n; i++) cout << ff[i] << ' ';
cout << endl;
}
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
#ifdef ACM_LOCAL
freopen("input", "r", stdin);
freopen("output", "w", stdout);
signed test_index_for_debug = 1;
char acm_local_for_debug = 0;
do {
if (acm_local_for_debug == '$') exit(0);
if (test_index_for_debug > 20)
throw runtime_error("Check the stdin!!!");
auto start_clock_for_debug = clock();
solve();
auto end_clock_for_debug = clock();
cout << "Test " << test_index_for_debug << " successful" << endl;
cerr << "Test " << test_index_for_debug++ << " Run Time: "
<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
cout << "--------------------------------------------------" << endl;
} while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
#else
solve();
#endif
return 0;
}
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