标签：style color io os ar for sp div on
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 si is not smaller than the age of person ti. 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.
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 si and ti. There is a blank line between every two cases. Process to the end of input.
For each the case, print the minimum number of groups that meet the requirement one line.
4 4 1 2 1 3 2 4 3 4
3
set1= {1}, set2= {2, 3}, set3= {4}
#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;
}
ZOJ 3795 Grouping(Tarjan缩点+DAG）
标签：style color io os ar for sp div on
原文地址：http://blog.csdn.net/u013582254/article/details/40252545
