Description
An annual bicycle rally will soon begin in Byteburg. The bikers of Byteburg are natural long distance cyclists. Local representatives of motorcyclists, long feuding the cyclists, have decided to sabotage the event.
There are intersections in Byteburg, connected with one way streets. Strangely enough, there are no cycles in the street network - if one can ride from intersection U to intersection V , then it is definitely impossible to get from V to U.
The rally‘s route will lead through Byteburg‘s streets. The motorcyclists plan to ride their blazing machines in the early morning of the rally day to one intersection and completely block it. The cyclists‘ association will then of course determine an alternative route but it could happen that this new route will be relatively short, and the cyclists will thus be unable to exhibit their remarkable endurance. Clearly, this is the motorcyclists‘ plan - they intend to block such an intersection that the longest route that does not pass through it is as short as possible.
给定一个N个点M条边的有向无环图，每条边长度都是1。
请找到一个点，使得删掉这个点后剩余的图中的最长路径最短。
Input
In the first line of the standard input, there are two integers, N and M(2<=N<=500 000,1<=M<=1 000 000), separated by a single space, that specify the number of intersections and streets in Byteburg. The intersections are numbered from to . The lines that follow describe the street network: in the -th of these lines, there are two integers, Ai, Bi(1<=Ai,Bi<=N,Ai<>Bi), separated by a single space, that signify that there is a one way street from the intersection no. Ai to the one no. Bi.
第一行包含两个正整数N,M(2<=N<=500 000,1<=M<=1 000 000)，表示点数、边数。
接下来M行每行包含两个正整数A[i],B[i] (1<=A[i],B[i]<=N,A[i]<>B[i])，表示A[i]到B[i]有一条边。
Output
The first and only line of the standard output should contain two integers separated by a single space. The first of these should be the number of the intersection that the motorcyclists should block, and the second - the maximum number of streets that the cyclists can then ride along in their rally. If there are many solutions, your program can choose one of them arbitrarily.
包含一行两个整数x,y，用一个空格隔开，x为要删去的点，y为删除x后图中的最长路径的长度，如果有多组解请输出任意一组。
Sample Input
6 5
1 3
1 4
3 6
3 4
4 5
Sample Output
1 2
Solution
一道神题（而且好像BZOJ上这段时间没开SPJ）
2:由于新换judge程序存在一些问题，致使OJ需要Spj的题提交失效，请不要提交此类试题，Bug正在排查中
这题没用主席树，但用了权值线段树
建一个源点和汇点
拓扑排序后，用dp的方法求得图上正向边的最长路\(d[0]\)数组和反向边的最长路\(d[1]\)数组，类似于SPFA的\(d\)数组，但不要用SPFA求。本来我用的SPFA，结果T掉了
那么对于每一条边，一定包含这条边的图上的最长路就是这条边的出发点的\(d[0]\)加上这条边到达点的\(d[1]\)，我们把这个值当做这条边的权值
那么删去一个点的话，就把以这个点为到达点的边的权值在权值线段树里删掉，然后就维护了删去了这个点后的最长路
更新答案后再把以这个点为出发点的边的权值加到权值线段树里
每次删掉一个点的时候并没有把与它相连的所有边都删掉，这样省时间
PS：这题还有一些地方没理解透，以后还会填坑
#include<bits/stdc++.h>
#define ll long long
#define db double
#define ld long double
#define Mid ((l+r)>>1)
#define lson rt<<1,l,Mid
#define rson rt<<1|1,Mid+1,r
const int MAXM=2000000+10,MAXN=2000000+10,inf=0x3f3f3f3f;
int n,m,e[2],beg[2][MAXN],nex[2][MAXM],to[2][MAXM],w[2][MAXM],s,t,d[2][MAXN],p[MAXN],degree[MAXN],ans=inf,num,topo[MAXN],cnt;
std::queue<int> q;
struct Q_Tree{
int Max[MAXM],Num[MAXM];
inline void PushUp(int rt)
{
if(Max[rt<<1]>Max[rt<<1|1])Max[rt]=Max[rt<<1],Num[rt]=Num[rt<<1];
else Max[rt]=Max[rt<<1|1],Num[rt]=Num[rt<<1|1];
}
inline void Insert(int rt,int l,int r,int pos)
{
if(l==r)Max[rt]=pos,Num[rt]++;
else
{
if(pos<=Mid)Insert(lson,pos);
else Insert(rson,pos);
PushUp(rt);
}
}
inline void Delete(int rt,int l,int r,int pos)
{
if(l==r)
{
Num[rt]--;
if(!Num[rt])Max[rt]=0;
}
else
{
if(pos<=Mid)Delete(lson,pos);
else Delete(rson,pos);
PushUp(rt);
}
}
};
Q_Tree T;
template<typename T> inline void read(T &x)
{
T data=0,w=1;
char ch=0;
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')w=-1,ch=getchar();
while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
x=data*w;
}
template<typename T> inline void write(T x,char c='\0')
{
if(x<0)putchar('-'),x=-x;
if(x>9)write(x/10);
putchar(x%10+'0');
if(c!='\0')putchar(c);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline void insert(int x,int y,int z)
{
to[0][++e[0]]=y;
nex[0][e[0]]=beg[0][x];
beg[0][x]=e[0];
w[0][e[0]]=z;
to[1][++e[1]]=x;
nex[1][e[1]]=beg[1][y];
beg[1][y]=e[1];
w[1][e[1]]=z;
}
inline void toposort()
{
for(register int i=1;i<=n;++i)
if(!degree[i])q.push(i);
while(!q.empty())
{
int x=q.front();
q.pop();
topo[++cnt]=x;
for(register int i=beg[0][x];i;i=nex[0][i])
{
degree[to[0][i]]--;
if(!degree[to[0][i]])q.push(to[0][i]);
}
}
}
int main()
{
read(n);read(m);
s=n+1;t=n+2;
for(register int i=1;i<=m;++i)
{
int u,v;
read(u);read(v);
degree[v]++;
insert(u,v,1);
}
toposort();
for(register int p=n;p>=1;--p)
for(register int x=topo[p],i=beg[0][x];i;i=nex[0][i])chkmax(d[1][x],d[1][to[0][i]]+1);
for(register int p=1;p<=n;++p)
for(register int x=topo[p],i=beg[0][x];i;i=nex[0][i])chkmax(d[0][to[0][i]],d[0][x]+1);
for(register int i=1;i<=n;++i)insert(s,i,0),insert(i,t,0);
for(register int i=1;i<=n;++i)T.Insert(1,1,n+2,d[1][i]);
d[0][s]=d[1][t]=-1;
for(register int t=1;t<=n;++t)
{
int x=topo[t];
for(register int i=beg[1][x];i;i=nex[1][i])T.Delete(1,1,n+2,d[0][to[1][i]]+d[1][x]+1);
if(T.Max[1]<ans)ans=T.Max[1],num=x;
for(register int i=beg[0][x];i;i=nex[0][i])T.Insert(1,1,n+2,d[0][x]+d[1][to[0][i]]+1);
}
write(num,' '),write(ans,'\n');
return 0;
}
