ZOJ 3795 Grouping(Tarjan缩点+DAG）

Suppose there are N people in ZJU, whose ages are unknown. We have some messages about them. The i-th message shows that the age of person s_i is not smaller than the age of person t_i. Now we need to divide all these N people into several groups. One's age shouldn't be compared with each other in the same group, directly or indirectly. And everyone should be assigned to one and only one group. The task is to calculate the minimum number of groups that meet the requirement.

Input

There are multiple test cases. For each test case: The first line contains two integers N(1≤ N≤ 100000), M(1≤ M≤ 300000), N is the number of people, and M is is the number of messages. Then followed by M lines, each line contain two integers s_i and t_i. There is a blank line between every two cases. Process to the end of input.

Output

For each the case, print the minimum number of groups that meet the requirement one line.

Sample Input

4 41 21 32 43 4

Sample Output

3

Hint

set1= {1}, set2= {2, 3}, set3= {4}

记忆化搜索最长路径：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<limits.h>typedef long long LL;using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int maxn=100100;const int maxm=300100;struct node{    int u,v;    int next;}e[maxm],e2[maxm];int head[maxn],cntE,cntF;int DFN[maxn],low[maxn],h[maxn];int s[maxm],top,dex,cnt;int belong[maxn],instack[maxn];int dp[maxn],num[maxn];int n,m;void init(){    top=cntE=cntF=0;    dex=cnt=0;    CLEAR(DFN,0);    CLEAR(head,-1);    CLEAR(instack,0);    CLEAR(num,0);//fuck num没清0wa了2小时}void addedge(int u,int v){    e[cntE].u=u;e[cntE].v=v;    e[cntE].next=head[u];    head[u]=cntE++;}void Tarjan(int u){    DFN[u]=low[u]=++dex;    instack[u]=1;    s[top++]=u;    for(int i=head[u];i!=-1;i=e[i].next)    {        int v=e[i].v;        if(!DFN[v])        {            Tarjan(v);            low[u]=min(low[u],low[v]);        }        else if(instack[v])            low[u]=min(low[u],DFN[v]);    }    int v;    if(DFN[u]==low[u])    {        cnt++;        do{            v=s[--top];            belong[v]=cnt;            instack[v]=0;        }while(u!=v);    }}int dfs(int x){    if(dp[x]) return dp[x];    dp[x]=num[x];    for(int i=h[x];i!=-1;i=e2[i].next)        dp[x]=max(dp[x],dfs(e2[i].v)+num[x]);    return dp[x];}void work(){    REPF(i,1,n)      if(!DFN[i])  Tarjan(i);    REPF(i,1,n)       num[belong[i]]++;    CLEAR(h,-1);    CLEAR(dp,0);    REPF(k,1,n)    {        for(int i=head[k];i!=-1;i=e[i].next)        {            int v=e[i].v;            if(belong[k]!=belong[v])            {                e2[cntF].u=belong[k];                e2[cntF].v=belong[v];                e2[cntF].next=h[belong[k]];                h[belong[k]]=cntF++;            }        }    }    int ans=0;//    cout<<"2333  "<<cnt<<endl;    REPF(i,1,cnt)        ans=max(ans,dfs(i));    printf("%d\n",ans);}int main(){    int u,v;    while(~scanf("%d%d",&n,&m))    {        init();        for(int i=0;i<m;i++)        {            scanf("%d%d",&u,&v);            addedge(u,v);        }        work();    }    return 0;}