标签：nes empty sequence drawing john 输出 path names else
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.
7 10 1 2 3 1 2 5 2 4 3 7 3 5 3 6 6 5 7 2 4 7
6
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.
先tarjan缩点，然后建边。在这个时候我们需要建两条边，一条跑正向，一条反向，说白了就是建一条反向边。
然后在进行双向dfs，跑出从1点所能到达的点的最长链及能到达一点的最长链，更新到达当前点时所能更新出的最大值。
然后进行枚举每一条边，我们将其进行反向，看其反向后是否能连通整个图，若能，更新最大值。
我们更新结果的时候，枚举每一条可以反向的边，只有在这条边可以从1出来并且可以回到1时才可以使用。
#include<queue> #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define N 210000 using namespace std; int n,m,x,y,s,tot,tat,top,tim; bool vis[N],vis1[N],vis2[N],vist1[N],vist2[N]; int xx[N],yy[N],in1[N],in2[N],head1[N],head2[N],dfn[N]; int low[N],sum[N],ans1[N],ans2[N],head[N],stack[N],belong[N]; queue<int>q; int read() { int x=0,f=1; char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1; ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘; ch=getchar();} return x*f; } struct Edge { int to,dis,next,from; }edge[N],edge1[N],edge2[N]; int add(int x,int y) { tot++; edge[tot].to=y; edge[tot].next=head[x]; head[x]=tot; } int add1(int x,int y) { tat++; edge1[tat].to=y; edge1[tat].next=head1[x]; edge2[tat].to=x; edge2[tat].next=head2[y]; head1[x]=head2[y]=tat; } int tarjan(int now) { dfn[now]=low[now]=++tim; vis[now]=true; stack[++top]=now; for(int i=head[now];i;i=edge[i].next) { int t=edge[i].to; if(vis[t]) low[now]=min(low[now],dfn[t]); else if(!dfn[t]) tarjan(t),low[now]=min(low[now],low[t]); } if(low[now]==dfn[now]) { s++,belong[now]=s,sum[s]++; for(;stack[top]!=now;top--) belong[stack[top]]=s,vis[stack[top]]=false,sum[s]++; vis[now]=false,top--; } } int shink_point() { for(int i=1;i<=n;i++) for(int j=head[i];j;j=edge[j].next) if(belong[i]!=belong[edge[j].to]) add1(belong[i],belong[edge[j].to]); } int dfs1(int x) { vis1[x]=true;vist1[x]=true; for(int i=head1[x];i;i=edge1[i].next) { int t=edge1[i].to; if(ans1[t]<ans1[x]+sum[t]) { ans1[t]=ans1[x]+sum[t]; dfs1(t); } } vis1[x]=false; } int dfs2(int x) { vis2[x]=true;vist2[x]=true; for(int i=head2[x];i;i=edge2[i].next) { int t=edge2[i].to; if(ans2[t]<ans2[x]+sum[t]) { ans2[t]=ans2[x]+sum[t]; dfs2(t); } } vis2[x]=false; } int main() { n=read(),m=read(); for(int i=1;i<=m;i++) xx[i]=read(),yy[i]=read(),add(xx[i],yy[i]); for(int i=1;i<=n;i++) if(!dfn[i]) tarjan(i); shink_point(); ans1[belong[1]]=ans2[belong[1]]=sum[belong[1]]; dfs1(belong[1]);dfs2(belong[1]); int answer=2*sum[belong[1]]; for(int i=1;i<=m;i++) { x=belong[yy[i]],y=belong[xx[i]]; if(vist1[x]&&vist2[y]) answer=max(answer,ans1[x]+ans2[y]); } printf("%d",answer-sum[belong[1]]); return 0; }
拓扑排序：
这个题原来是打算用来练拓扑排序的，结果做了一天的拓扑排序发现不过样例、、、、
为什么会用拓扑排序？？
因为我们用拓扑排序的话可以轻易地找到最长链。
怎么拓扑排序？？
我们如果直接进行拓扑排序的话，我们会意识到一个问题：缩完点以后直接统计出来入度为零的点并非是我们所需要的点1，我们要跑最长链的话我们需要从1点开始跑，也就是说我们的起点必须是1，怎样做到这一点？？我们要做到起点是一的话我们必须让1的入度为零，从一点开始更新与他相连的点。从新统计他们的入读，也就是说我们将这个可能出现环的图抽离成一颗树，这棵树的树根为1点。然后再进行拓扑排序，找出最长链。
最后在进行枚举边，进行更新、
#include<queue> #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define N 210000 using namespace std; int n,m,x,y,s,tot,tat,top,tim; bool vis[N],vis1[N],vis2[N]; int xx[N],yy[N],in1[N],in2[N],head1[N],head2[N],dfn[N]; int low[N],sum[N],ans1[N],ans2[N],head[N],stack[N],belong[N]; int read() { int x=0,f=1; char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1; ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘; ch=getchar();} return x*f; } struct Edge { int to,from,next; }edge[N],edge1[N],edge2[N]; int add(int x,int y) { tot++; edge[tot].to=y; edge[tot].next=head[x]; head[x]=tot; } int add1(int x,int y) { tat++; edge1[tat].to=y; edge1[tat].next=head1[x]; edge2[tat].to=x; edge2[tat].next=head2[y]; head1[x]=head2[y]=tat; } int tarjan(int now) { dfn[now]=low[now]=++tim; vis[now]=true;stack[++top]=now; for(int i=head[now];i;i=edge[i].next) { int t=edge[i].to; if(vis[t]) low[now]=min(dfn[t],low[now]); else if(!dfn[t]) tarjan(t),low[now]=min(low[t],low[now]); } if(low[now]==dfn[now]) { s++,belong[now]=s,sum[s]++; for(;stack[top]!=now;top--) belong[stack[top]]=s,sum[s]++,vis[stack[top]]=false; vis[now]=false;top--; } } int shink_point() { for(int i=1;i<=m;i++) for(int j=head[i];j;j=edge[j].next) if(belong[i]!=belong[edge[j].to]) add1(belong[i],belong[edge[j].to]); } int dfs1(int s) { for(int i=head1[s];i;i=edge1[i].next) { int t=edge1[i].to; if(!in1[t]) dfs1(t); in1[t]++; } } int dfs2(int s) { for(int i=head2[s];i;i=edge2[i].next) { int t=edge2[i].to; if(!in2[t]) dfs2(t); in2[t]++; } } int tpsort(int *in,Edge *edge,int *head,bool *vis,int *ans) { queue<int>q; q.push(belong[1]); while(!q.empty()) { int x=q.front();q.pop();vis[x]=true; for(int i=head[x];i;i=edge[i].next) { int t=edge[i].to; in[t]--; if(!in[t]) q.push(t); ans[t]=max(ans[t],ans[x]+sum[t]); } } } int main() { n=read(),m=read(); int answer=0; for(int i=1;i<=m;i++) xx[i]=read(),yy[i]=read(),add(xx[i],yy[i]); for(int i=1;i<=n;i++) if(!dfn[i]) tarjan(i); shink_point(); dfs1(belong[1]),dfs2(belong[1]); ans1[belong[1]]=ans2[belong[1]]=sum[belong[1]]; tpsort(in1,edge1,head1,vis1,ans1); tpsort(in2,edge2,head2,vis2,ans2); answer=2*sum[belong[1]]; for(int i=1;i<=m;i++) { x=belong[yy[i]],y=belong[xx[i]]; if(vis1[x]&&vis2[y]) answer=max(answer,ans1[x]+ans2[y]); } printf("%d",answer-sum[belong[1]]); return 0; }
洛谷——P3119 [USACO15JAN]草鉴定Grass Cownoisseur
标签：nes empty sequence drawing john 输出 path names else
原文地址：http://www.cnblogs.com/z360/p/7471352.html
