洛谷P3119 USACO15JAN 草鉴定
题目描述
In an effort to better manage the grazing patterns of his cows, Farmer John has installed one-way cow paths all over his farm. The farm consists of N fields, conveniently numbered 1..N, with each one-way cow path connecting a pair of fields. For example, if a path connects from field X to field Y, then cows are allowed to travel from X to Y but not from Y to X.
Bessie the cow, as we all know, enjoys eating grass from as many fields as possible. She always starts in field 1 at the beginning of the day and visits a sequence of fields, returning to field 1 at the end of the day. She tries to maximize the number of distinct fields along her route, since she gets to eat the grass in each one (if she visits a field multiple times, she only eats the grass there once).
As one might imagine, Bessie is not particularly happy about the one-way restriction on FJ's paths, since this will likely reduce the number of distinct fields she can possibly visit along her daily route. She wonders how much grass she will be able to eat if she breaks the rules and follows up to one path in the wrong direction. Please compute the maximum number of distinct fields she can visit along a route starting and ending at field 1, where she can follow up to one path along the route in the wrong direction. Bessie can only travel backwards at most once in her journey. In particular, she cannot even take the same path backwards twice.
约翰有n块草场，编号1到n，这些草场由若干条单行道相连。奶牛贝西是美味牧草的鉴赏家，她想到达尽可能多的草场去品尝牧草。
贝西总是从1号草场出发，最后回到1号草场。她想经过尽可能多的草场，贝西在通一个草场只吃一次草，所以一个草场可以经过多次。因为草场是单行道连接，这给贝西的品鉴工作带来了很大的不便，贝西想偷偷逆向行走一次，但最多只能有一次逆行。问，贝西最多能吃到多少个草场的牧草。
这道题还是很复杂的。
根据去品尝牧草，我们很容易想到，如果一些草场在一个连通分量里面，只要到达这个连通分量，里面所有的草场就能到达，所以Tarjan缩点是无疑的了。
缩完点以后，我们保证不存在环，这样就可以跑一遍最长路，其中边权是每个连通分量的大小。
但是，还有能倒着走一条边的情况。这里就需要用到分层最短路。
在对于缩完点的图中，边u->v，我们将v连向u+n，代表下一层，当然u+n也要连到v+n，这样，我们能从每一个点，倒着走到下一层，但是无法从下一层走回来。
然后再输出起点连通分量+n的最长路dis值就好了。
但是注意，如果整个图缩完点是一个连通分量，那么要输出起点连通分量，不用加n。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cstdlib> #include <stack> #include <queue> #define REP(i,k,n) for(int i=k;i<=n;i++) #define in(a) a=read() #define MAXN 200010 using namespace std; inline int read(){ int x=0,f=1; char ch=getchar(); for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-1; for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0'; return x*f; } stack <int> S; queue <int> Q; int n,m; int total1,head1[MAXN],to1[MAXN],nxt1[MAXN]; int total2,head2[MAXN<<2],to2[MAXN<<2],nxt2[MAXN<<2],val[MAXN<<2]; int num,cnt,dfn[MAXN],low[MAXN],vis[MAXN],bel[MAXN],size[MAXN],dis[MAXN]; inline void adl1(int a,int b){ total1++; to1[total1]=b; nxt1[total1]=head1[a]; head1[a]=total1; return ; } inline void adl2(int a,int b,int c){ total2++; to2[total2]=b; val[total2]=c; nxt2[total2]=head2[a]; head2[a]=total2; return ; } inline void tarjan(int u){ dfn[u]=low[u]=++cnt; S.push(u),vis[u]=1; for(int e=head1[u];e;e=nxt1[e]){ if(!dfn[to1[e]]){ tarjan(to1[e]); low[u]=min(low[u],low[to1[e]]); } else if(vis[to1[e]]) low[u]=min(low[u],dfn[to1[e]]); } if(dfn[u]==low[u]){ num++; while(!S.empty() && S.top()!=u) size[num]++,bel[S.top()]=num,vis[S.top()]=0,S.pop(); if(!S.empty()) size[num]++,bel[S.top()]=num,vis[S.top()]=0,S.pop(); } return ; } inline void spfa(){ Q.push(bel[1]); dis[bel[1]]=0; while(!Q.empty()){ int u=Q.front(); Q.pop(),vis[u]=0; for(int e=head2[u];e;e=nxt2[e]) if(dis[to2[e]]<dis[u]+val[e]){ dis[to2[e]]=dis[u]+val[e]; if(!vis[to2[e]]) vis[to2[e]]=1,Q.push(to2[e]); } } return ; } int main(){ in(n),in(m); int a,b; REP(i,1,m) in(a),in(b),adl1(a,b); REP(i,1,n) if(!dfn[i]) tarjan(i); REP(u,1,n) for(int e=head1[u];e;e=nxt1[e]) if(bel[u]!=bel[to1[e]]){ adl2(bel[u],bel[to1[e]],size[bel[u]]); adl2(bel[u]+num,bel[to1[e]]+num,size[bel[u]]); adl2(bel[to1[e]],bel[u]+num,size[bel[to1[e]]]); } adl2(bel[1],bel[1]+num,size[bel[1]]); memset(vis,0,sizeof(vis)); spfa(); cout<<dis[bel[1]+num]<<endl; return 0; }
