Andrew and Taxi CodeForces - 1100E (思维,拓扑)
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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<<‘ ‘ #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 ‘ ‘ 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 ",ans,int(path.size())); for (int i:path) printf("%d ",i);hr; }
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