题目链接

BZOJ3832

题解

神思路orz，根本不会做

设\(f[i]\)为到\(i\)的最长路，\(g[i]\)为\(i\)出发的最长路，二者可以拓扑序后\(dp\)求得
那么一条边\((u,v)\)的对应的最长链就是\(f[u] + 1 + g[v]\)
我们人为加入源汇点\(S\)，\(T\)，\(S\)向每个点连边，每个点向\(T\)连边
我们考虑把整个图划分开
一开始所有点都在\(T\)这边，割边为所有\(S\)的边
然后我们按照拓扑序把点逐一加入\(S\)集合中
加入时，我们删去\(S\)集合连向该点的边，然后询问所有边的最大值，即为删去该点的最长链
加入后，我们加入该点连向\(T\)集合的边
由于是按照拓扑序，所以以上提到的所有边就是该点的所有入边/出边

然后所有边的最大值可以用堆或者线段树维护

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u]; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 500005,maxm = 2000005,INF = 1000000000;
inline int read(){int out = 0,flag = 1; char c = getchar();while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}return out * flag;
}
struct Heap{priority_queue<int> a,b;void ck(){while (!b.empty() && a.top() == b.top()) a.pop(),b.pop();}int size(){return a.size() - b.size();}void ins(int x){ck(); a.push(x);}void del(int x){ck(); b.push(x);}int top(){ck(); return size() ? a.top() : 0;}
}H;
int n,m,f[maxn],g[maxn],s[maxn];
int q[maxn],head,tail;
int h[maxn],de[maxn],ne;
int hi[maxn],nei;
struct EDGE{int to,nxt;}ed[maxm],e[maxm];
inline void build(int u,int v){ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;de[v]++;e[++nei] = (EDGE){u,hi[v]}; hi[v] = nei;
}
void topu(){head = 0; tail = -1; int u;REP(i,n) if (!de[i]) q[++tail] = i;REP(i,n){s[i] = u = q[head++];Redge(u) if (!(--de[ed[k].to])) q[++tail] = ed[k].to;}
}
void init(){REP(i,n){int u = s[i];Redge(u) f[ed[k].to] = max(f[ed[k].to],f[u] + 1);}for (int i = n; i; i--){int u = s[i];Redge(u) g[u] = max(g[u],g[ed[k].to] + 1);}
}
void solve(){int ans = INF,ansu = 0,x;REP(i,n) H.ins(g[i]);REP(i,n){int u = s[i];H.del(g[u]);for (int k = hi[u]; k; k = e[k].nxt)H.del(f[e[k].to] + 1 + g[u]);x = H.top();if (x < ans) ans = x,ansu = u;H.ins(f[u]);Redge(u) H.ins(f[u] + 1 + g[ed[k].to]);}printf("%d %d\n",ansu,ans);
}
int main(){n = read(); m = read();int a,b;REP(i,m){a = read(); b = read();build(a,b);}topu();init();//REP(i,n) printf("node%d  f = %d   g = %d\n",i,f[i],g[i]);solve();return 0;
}