ZOJ 3795 Grouping（强连通+最长路）

题意：有n个人，m个关系，每个关系表示si的体重>=ti的体重，问你最少划分多少个组，使得这些组里面的人不能直接或者间接和组里的人比体重

思路：首先要考虑到一个问题，si>=ti，那么就可能存在几个人的体重都是一样的，那么他们几个人就是等价的，所以需要先缩点，然后就显然的转化成求DAG上的最长路了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
const int maxn = 100000+7;
vector<int>e[maxn];
int pre[maxn],lowlink[maxn],sccno[maxn],dfs_clock,scc_cnt;
int cnt[maxn];
stack<int>s;
void dfs(int u)
{
pre[u]=lowlink[u]=++dfs_clock;
s.push(u);
for(int i = 0;i<e[u].size();i++)
{
int v = e[u][i];
if(!pre[v])
{
dfs(v);
lowlink[u]=min(lowlink[u],lowlink[v]);
}
else if(!sccno[v])
lowlink[u]=min(lowlink[u],pre[v]);
}
if(lowlink[u]==pre[u])
{
scc_cnt++;
for(;;)
{
int x = s.top();s.pop();
sccno[x]=scc_cnt;
cnt[scc_cnt]++;
if(x==u)break;
}
}
}
void find_scc(int n)
{
dfs_clock=scc_cnt=0;
memset(sccno,0,sizeof(sccno));
memset(cnt,0,sizeof(cnt));
memset(pre,0,sizeof(pre));
while(!s.empty())s.pop();
for(int i = 1;i<=n;i++)
if(!pre[i])
dfs(i);
}
int in[maxn];
vector<int>G[maxn];
int ans = 0;
int dp[maxn];
void bfs()
{
queue<int>q;
memset(dp,0,sizeof(dp));
for(int i = 1;i<=scc_cnt;i++)
if(!in[i])
{
q.push(i);
dp[i]=cnt[i];
}
while(!q.empty())
{
int u = q.front();q.pop();
for(int i =0;i<G[u].size();i++)
{
int v = G[u][i];
dp[v]=max(dp[u]+cnt[v],dp[v]);
if(--in[v]==0)q.push(v);
}
}
for(int i = 1;i<=scc_cnt;i++)
ans=max(ans,dp[i]);
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
ans = 0;
memset(in,0,sizeof(in));
for(int i = 0;i<=n;i++)e[i].clear();
for(int i = 0;i<=n;i++)G[i].clear();
for(int i = 1;i<=m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
e[u].push_back(v);
}
find_scc(n);
for(int u=1;u<=n;u++)
{
for(int i = 0;i<e[u].size();i++)
{
int v = e[u][i];
if(sccno[u]!=sccno[v])
{
in[sccno[v]]++;
G[sccno[u]].push_back(sccno[v]);
}
}
}
bfs();
printf("%d\n",ans);
}
}


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.

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 si and ti. 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 4
1 2
1 3
2 4
3 4

Sample Output

3

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