Andrew and Taxi CodeForces - 1100E (思维,拓扑)

时间：2022-07-01 01:15:48

大意: 给定有向图, 每条边有一个权值, 假设你有$x$个控制器, 那么可以将所有权值不超过$x$的边翻转, 求最少的控制器数, 使得翻转后图无环

先二分转为判定问题. 每次check删除能动的边, 若剩余图有环显然不成立, 否则将剩余的图拓排一下, 再把能动的边按拓排的方向即可保证无环.

#include <iostream>

#include <algorithm>

#include <math.h>

#include <cstdio>

#include <set>

#include <map>

#include <string>

#include <vector>

#include <string.h>

#include <queue>

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define hr cout<<'\n'

#define pb push_back

#define mid ((l+r)>>1)

#define lc (o<<1)

#define rc (lc|1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false);

#define endl '\n'

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+7;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}

void exgcd(ll a,ll b,ll &d,ll &x,ll &y){b?exgcd(b,a%b,d,y,x),y-=a/b*x:x=1,y=0,d=a;}

ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}

//head

const int N = 4e5+10, INF = 0x3f3f3f3f;

int n, m;

struct _ {int u,v,w;}e[N];

vector<int> g[N];

int deg[N], s[N];

int chk(int x) {

	REP(i,1,n) g[i].clear(),deg[i]=0;

	REP(i,1,m) {

		if (e[i].w>x) {

			g[e[i].u].pb(e[i].v);

			++deg[e[i].v];

		}

	}

	queue<int> q;

	*s = 0;

	REP(i,1,n) if (!deg[i]) q.push(i),s[i]=++*s;

	while (!q.empty()) {

		int u = q.front();

		q.pop();

		for (int v:g[u]) {

			if (!--deg[v]) q.push(v),s[v]=++*s;

		}

	}

	REP(i,1,n) if (deg[i]) return 0;

	return 1;

}

int main() {

	scanf("%d%d", &n, &m);

	REP(i,1,m) scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);

	int l = 0, r = 1e9, ans;

	while (l<=r) {

		if (chk(mid)) ans=mid,r=mid-1;

		else l=mid+1;

	}

	chk(ans);

	vector<int> path;

	REP(i,1,m) {

		if (e[i].w<=ans&&s[e[i].u]>s[e[i].v]) {

			path.pb(i);

		}

	}

	printf("%d %d\n",ans,int(path.size()));

	for (int i:path) printf("%d ",i);hr;

}

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