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。
输入格式
INPUT: (file grass.in)
The first line of input contains N and M, giving the number of fields and the number of one-way paths (1 <= N, M <= 100,000).
The following M lines each describe a one-way cow path. Each line contains two distinct field numbers X and Y, corresponding to a cow path from X to Y. The same cow path will never appear more than once.
输入：
第一行：草场数n，道路数m。
以下m行，每行x和y表明有x到y的单向边，不会有重复的道路出现。
输出格式
OUTPUT: (file grass.out)
A single line indicating the maximum number of distinct fields Bessie
can visit along a route starting and ending at field 1, given that she can
follow at most one path along this route in the wrong direction.
输出：
一个数，逆行一次最多可以走几个草场。
输入输出样例
7 10 1 2 3 1 2 5 2 4 3 7 3 5 3 6 6 5 7 2 4 7
6
说明/提示
SOLUTION NOTES:
Here is an ASCII drawing of the sample input:
v---3-->6
7 | |
^ v |
| 1 |
| | v
| v 5
4<--2---^
Bessie can visit pastures 1, 2, 4, 7, 2, 5, 3, 1 by traveling
backwards on the path between 5 and 3. When she arrives at 3 she
cannot reach 6 without following another backwards path.
思路
![P3119 [USACO15JAN]Grass Cownoisseur G [ Tarjan + 缩点 + 拓扑序 + dp + 最长路] [好题]第1张](https://img2018.cnblogs.com/i-beta/1488084/202002/1488084-20200229014601022-1260118161.png)
相当坎坷的一道紫题,数据也不算水了, 前前后后大概想了两个小时, 写了一个小时, wa了一个小时.
首先, 知道这题是求最长路;
其次, 注意到反向只能走一次这个问题, 怎么做到只能反向走一次.
先把点分成三种情况来讨论一下,
1 直接可以由 1 号草坪走来的;
2 可以走到 1 号草坪的;
3 和 1 号点完全没有任何关系的.
显然这道题根本不会用到第 3 这种情况, 所以只需要建立一个正图和一个反向图, 在DAG上跑最长路, 显然可以由 1 号节点做起点在拓扑序上跑dp就可以了.
最后枚举每条反边, ans 就是 反向走到此条反边的正向图的父亲边的距离 + 正向走到此条反边的距离 - 1 号节点所在的连通块大小( 就是这里wa穿了, 因为有数据是1号节点直接在一个连通块里的 ), 并维护 ans 的最大值即可.
CODE
![P3119 [USACO15JAN]Grass Cownoisseur G [ Tarjan + 缩点 + 拓扑序 + dp + 最长路] [好题]第2张](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
![P3119 [USACO15JAN]Grass Cownoisseur G [ Tarjan + 缩点 + 拓扑序 + dp + 最长路] [好题]第3张](https://images.cnblogs.com/OutliningIndicators/ExpandedBlockStart.gif)
1 #include <bits/stdc++.h> 2 #define dbg(x) cout << #x << "=" << x << endl 3 4 using namespace std; 5 typedef long long LL; 6 const int maxn = 1e5 + 7; 7 8 int head[maxn], dfn[maxn], low[maxn], st[maxn]; 9 int cnt = 0, tot = 0, tim = 0, top = 1, n, m, cl = 0, ans = 0; 10 int vis[maxn]; 11 int color[maxn]; 12 int sz[maxn]; 13 int dis[maxn][5]; 14 int head1[maxn << 1][5], cnt1, edge1[maxn << 1][5], nxt1[maxn << 1][5]; 15 int in[maxn << 1][5]; 16 17 /* 18 head[],结构体edge:存边 19 20 dfn[],low[]:tarjan中数组 21 22 st[]:模拟栈 23 24 out[]:出边 25 26 sd[]:强连通分量存储 27 28 dq[]:统计答案 29 */ 30 31 template<class T>inline void read(T &res) 32 { 33 char c;T flag=1; 34 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0'; 35 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag; 36 } 37 38 struct Edge{ 39 int nxt, to; 40 }edge[maxn * 2]; 41 42 inline void BuildGraph(int from, int to) 43 { 44 cnt++; 45 edge[cnt].to = to; 46 edge[cnt].nxt = head[from]; 47 head[from] = cnt; 48 } 49 50 void tarjan(int x) 51 { 52 tim++; 53 dfn[x] = low[x] = tim; 54 st[top] = x; 55 top++; 56 vis[x] = 1; 57 for(int i = head[x] ; i != 0; i = edge[i].nxt) 58 { 59 int u = edge[i].to; 60 if(vis[u] == 0) 61 { 62 tarjan(u); 63 low[x]=min(low[x],low[u]); 64 } 65 else if(vis[u] == 1) 66 low[x]=min(low[x],dfn[u]); 67 } 68 if(dfn[x] == low[x]) 69 { 70 cl++; 71 do 72 { 73 top--; 74 color[st[top]] = cl; 75 vis[st[top]] = -1; 76 sz[color[st[top]]]++; 77 }while( st[top] != x ); 78 } 79 return ; 80 } 81 82 void addedge(int u, int v, int cas) { 83 if(cas == 1) { 84 cnt++; 85 } 86 in[v][cas]++; 87 edge1[cnt][cas] = v; 88 nxt1[cnt][cas] = head1[u][cas]; 89 head1[u][cas] = cnt; 90 } 91 92 void topo(int cas) { 93 dis[color[1]][cas] = sz[color[1]]; 94 queue<int> q; 95 for ( int i = 1; i <= cl; ++i ) { 96 if(in[i][cas] == 0) { 97 q.push(i); 98 } 99 } 100 while(!q.empty()) { 101 int u = q.front(); 102 q.pop(); 103 for ( int i = head1[u][cas]; i; i = nxt1[i][cas] ) { 104 int v = edge1[i][cas]; 105 dis[v][cas] = max(dis[v][cas], dis[u][cas] + sz[v]); 106 if(--in[v][cas] == 0) { 107 q.push(v); 108 } 109 } 110 } 111 } 112 113 int main() 114 { 115 scanf("%d %d",&n, &m); 116 for ( int i = 1; i <= m; ++i ) { 117 int x, y; 118 scanf("%d %d",&x, &y); 119 BuildGraph(x, y); 120 } 121 for ( int i = 1; i <= n; ++i ) { 122 if( !vis[i] ) { 123 tarjan(i); 124 } 125 } 126 cnt = 0; 127 for ( int i = 1; i <= n; ++i ) { 128 for ( int j = head[i]; j; j = edge[j].nxt ) { 129 int v = edge[j].to; 130 if(color[i] != color[v]) { 131 addedge(color[i], color[v], 1); 132 addedge(color[v], color[i], 2); 133 } 134 } 135 } 136 memset(dis, 0xef, sizeof(dis)); 137 ans = sz[color[1]]; 138 topo(1), topo(2); 139 for ( int i = 1; i <= n; ++i ) { 140 for ( int j = head[i]; j; j = edge[j].nxt ) { 141 int v = edge[j].to; 142 if(color[i] != color[v]) { 143 ans = max(ans, dis[color[v]][1] + dis[color[i]][2] - sz[color[1]]); 144 } 145 } 146 } 147 cout << ans << endl; 148 return 0; 149 }