[Usaco15jan] Grass identification grass cownoisseur

"Title description":
[Usaco15jan] Grass identification grass cownoisseur

"Ideas":
First of all, let's consider that Bessie does not take the opposite edge, so for any strongly connected component \ (e\) , it is easy to know:

• \ (\forall u,v \in e\),\ (\exists u \to v \ and \ v \to u\)

\ (\because\) Bessie only eats grass once every time she passes through a pasture,\therefore\ can be shrunk, and then a \ (dag\)is obtained to count each of the strong connected components \ ( size\) value that indicates how many points are in this strongly connected component, and then run \ (dp\) on \ (dag\) .

But the problem is that Bessie plays snake skin, and she can be retrograde once.

So we should change a way of thinking, after shrinking points, considering the situation of retrograde, we can put the positive and negative of the chart is built once, set 1th points where the strong connected component is \ (e\), find out \ (dis1[]\) and \ (dis2[]\), Represents the longest road with \ (e\) as the starting point and the midpoint with \ (e\) , and then enumerates each \ (i\), then the answer is:

• \ (Max \{dis1[i] +dis2[v]-size[e]\}\).

This is the equivalent of taking the opposite side.

#include <cstdio> #include <algorithm> #include <cstring> #include <queue> #define INF 0x3f3f3f3fusing namespace Std;int n,m;const int maxn = 100005;struct edge{int u,v,nxt;} E[maxn];int head[maxn];int cnt = 0;int dfn[maxn];int low[maxn];int id = 0;int S[maxn];int tot = 0;int Bcnt = 0;int X[MAXN] ; int Y[maxn];bool vis[maxn];int size[maxn];int b[maxn];bool back[maxn];int dis1[maxn];int dis2[MAXN];int ans =-inf; inline void Add (int u,int v) {e[++cnt].u = U;E[CNT].V = V;E[CNT].NXT = Head[u];head[u] = cnt;}    inline void Tarjan (int u) {low[u] = dfn[u] = ++id;    S[++tot]=u;vis[u] = 1;        for (int i=head[u];i;i=e[i].nxt) {int v = E[I].V;            if (!dfn[v]) {Tarjan (v);        Low[u] = min (Low[u], low[v]);    } else if (Vis[v]) low[u] = min (Low[u], dfn[v]); } if (dfn[u] = = Low[u]) {int j=0;        bcnt++;            while (j ^ u) {j = s[tot--];            VIS[J] = 0;            B[J] = bcnt;        size[bcnt]++;  }  }}inline BOOL CMP (int a,int b) {return a > B;}    Queue<int>q;inline void dp (int *dis) {memset (dis,0,sizeof dis);    memset (vis,0,sizeof Vis);    DIS[B[1]] = size[b[1]];        Q.push (b[1]);        while (!q.empty ()) {int u = q.front (); Q.pop (); vis[u] = 0;            for (int i=head[u];i;i=e[i].nxt) {int v = E[I].V;                if (Dis[v] < Dis[u] + Size[v]) {Dis[v] = Dis[u] + size[v];                    if (!vis[v]) {Q.push (v);                VIS[V] = 1;    }}}}}int main () {scanf ("%d%d", &n,&m);        for (int i=1;i<=m;++i) {int u,v;scanf ("%d%d", &u,&v);        Add (U,V);    X[i] = u;y[i] = v;        } for (int i=1;i<=n;++i) {if (!dfn[i]) {Tarjan (i);    }} memset (head,0,sizeof head); cnt = 0;        for (int i=1;i<=m;++i) {if (B[x[i] ^ b[y[i]]) {Add (B[x[i]], b[y[i]]);   }} for (int i=head[b[1]];i;i=e[i].nxt) {     int v = E[I].V;    BACK[V] = 1;        } DP (DIS1);    memset (head,0,sizeof head); cnt = 0;    memset (back,0,sizeof back);        for (int i=1;i<=m;++i) {if (B[x[i] ^ b[y[i]]) {Add (B[y[i]], b[x[i]]);        }} for (int i=head[b[1]];i;i=e[i].nxt) {int v = E[I].V;    BACK[V] = 1;        } DP (DIS2);        for (int i=1;i<=bcnt;++i) {if (!dis1[i]) continue;            for (int j=head[i];j;j=e[j].nxt) {int v = E[J].V;            if (!dis2[v]) continue;        ans = max (ans, (dis1[i] + dis2[v]-size[b[1]));    }} printf ("%d", ans); return 0;}

[Usaco15jan] grass identification grass cownoisseur