uva 11324 The Largest Clique(强连通分量缩点+DAG动态规划)

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=25&page=show_problem&problem=2299

题意：输入n和m，有n个点和m条有向边，求出一个节点集合包括的节点个数最多，而且该节点内的不论什么两点a,b，要么a能到达b，要么b能到达a，要么a和b互相到达。

思路：强连通分量缩点形成有向无环图DAG，把缩点后的每一个点的权值置为该强连通分量的节点个数。最后在求DAG上的动态规划。

#include 
#include 
#include 
#include 
#include 
#define LL long long
#define _LL __int64

using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1010;

vector  edge[maxn],edge2[maxn];
int n,m;
int dfn[maxn],low[maxn],instack[maxn],dep,scc;
stack  st;
int set[maxn],num[maxn];
int d[maxn];

void init()
{
    for(int i = 1; i <= n; i++)
    {
        edge[i].clear();
        edge2[i].clear();
    }
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(instack,0,sizeof(instack));
    while(!st.empty()) st.pop();

    dep = 0;
    scc = 0;
    memset(num,0,sizeof(num));
    memset(d,0,sizeof(d));
}

void tarjan(int u)
{
    dfn[u] = low[u] = ++dep;
    instack[u] = 1;
    st.push(u);

    for(int i = 0; i < (int)edge[u].size(); i++)
    {
        int v = edge[u][i];
        if(!dfn[v])
        {
            tarjan(v);
            low[u] = min(low[u],low[v]);
        }
        else if(instack[v])
            low[u] = min(low[u],dfn[v]);
    }
    if(dfn[u] == low[u])
    {
        scc++;
        int t;
        while(1)
        {
            t = st.top();
            st.pop();
            instack[t] = 0;
            set[t] = scc;
            num[scc]++;
            if(t == u)
                break;
        }
    }
}

void creat()
{
    for(int u = 1; u <= n; u++)
    {
        for(int i = 0; i < (int)edge[u].size(); i++)
        {
            int v = edge[u][i];
            if(set[u] != set[v])
                edge2[set[u]].push_back(set[v]);
        }
    }
}

int dp(int u)
{
    if(d[u]) return d[u];
    else if(edge2[u].size() == 0) return d[u] = num[u];

    int ans = 0;
    for(int i = 0; i < (int)edge2[u].size(); i++)
    {
        int v = edge2[u][i];
        ans = max(ans,dp(v));
    }
    return d[u] = ans+num[u];
}

int main()
{
    int test,u,v;
    scanf("%d",&test);
    while(test--)
    {
        init();
        scanf("%d %d",&n,&m);
        for(int i = 1; i <= m; i++)
        {
            scanf("%d %d",&u,&v);
            if(u == v) continue;
            edge[u].push_back(v);
        }
        for(int i = 1; i <= n; i++)
            if(!dfn[i])
                tarjan(i);

        creat();

        int ans = 0;
        for(int i = 1; i <= scc; i++)
        {
            ans = max(ans,dp(i));
        }
        printf("%d\n",ans);
    }
    return 0;
}