Educational Codeforces Round 10 E. Pursuit For Artifacts(边双连通缩点)

题意:

$N,M\le 3\times 10^5,N个点,M条边的无向图,无重边自环$
$边权为0或1,问s\to t是否存权\ge 1且每条边经过一次的一条路径$

分析:

$边双缩点成树,树边(桥边)权保留,bcc内总权映射到点上$
$ans=dis(s,t)\ge 1$
$正确性就是bcc任意2点连通,是个简单环嘛,选有1的那半边走就好啦$ $并且树上2点路径是唯一的,满足题意就做完了$
$时间复杂度O(n+m)$

代码:

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//
// Created by TaoSama on 2016-03-25
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 3e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, m;
struct Edge {
int v, c;
};
vector<Edge> G[N], T[N];
int dfn[N], low[N], in[N], id[N], bcc, dfsNum;
int stk[N], top;
void tarjan(int u, int f) {
dfn[u] = low[u] = ++dfsNum;
stk[++top] = u;
in[u] = true;
for(Edge e : G[u]) {
int v = e.v;
if(v == f) continue;
if(!dfn[v]) {
tarjan(v, u);
low[u] = min(low[u], low[v]);
} else low[u] = min(low[u], dfn[v]);
}
if(low[u] == dfn[u]) {
++bcc;
while(true) {
int v = stk[top--];
in[v] = false;
id[v] = bcc;
if(v == u) break;
}
}
}
void init() {
bcc = dfsNum = 0;
memset(dfn, 0, sizeof dfn);
tarjan(1, -1);
}
int val[N];
bool dfs(int s, int t, int fa, int sum) {
if(s == t) return sum + val[t];
for(Edge e : T[s]) {
int v = e.v;
if(v == fa) continue;
if(dfs(v, t, s, sum + e.c + val[v])) return true;
}
return false;
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; ++i) {
int u, v, c; scanf("%d%d%d", &u, &v, &c);
G[u].push_back({v, c});
G[v].push_back({u, c});
}
init();
for(int i = 1; i <= n; ++i) {
int u = id[i];
for(Edge e : G[i]) {
int v = id[e.v];
if(u == v) val[u] += e.c;
else T[u].push_back({v, e.c});
}
}
int s, t; scanf("%d%d", &s, &t);
puts(dfs(id[s], id[t], -1, val[id[s]]) ? "YES" : "NO");
return 0;
}


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