Uva11324-the largest clique (DAG + dp + SCC)

Problem B: the largest clique

Given a Directed GraphG, Consider the following transformation. First, create a new GraphT (G)To have the same vertex setG. Create a directed edge between two verticesUAndVInT (G)If and only if there is a pathUAndVInGThat follows the directed edges only in the forward direction. This graphT (G)Is often calledTransitive ClosureOfG.

We defineCliqueIn a directed graph as a set of verticesUSuch that for any two verticesUAndVInU, There is a directed edge either fromUToVOr fromVToU(Or both). The size of a clique is the number of vertices in the clique.

The number of cases is given on the first line of input. Each test case describes a graphG. It begins with a line of two integersNAndM, Where 0 ≤N≤ 1000 is the number of verticesGAnd 0 ≤M≤ 50,000 is the number of directed edgesG. The verticesGAre numbered from 1N. The followingMLines contain two distinct integersUAndVBetween 1 andNWhich define a directed edge fromUToVInG.

For each test case, output a single integer that is the size of the largest clique inT (G).

Sample Input

15 51 22 33 14 15 2

Output for sample input

4

Strongly Connected Component shrinkage point + Dag DP

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <string>#include <algorithm>#include <queue>#include <stack>using namespace std;const int maxn = 1000+10;const int maxm = 50000+10;struct edge{    int v,nxt;}e[maxm];int head[maxn];int nume;stack<int> S;int dfs_clk;int scc_no;int dfn[maxn],lown[maxn],sccno[maxn];int dp[maxn];int n,m;int num[maxn];vector<int> G[maxn];void init(){    while(!S.empty()) S.pop();    memset(head,0,sizeof head);    memset(num,0,sizeof num);    memset(dp,-1,sizeof dp);    for(int i = 0; i <= n; i++){        G[i].clear();    }    nume = 1;    dfs_clk = 0;    scc_no = 0;    memset(dfn,0,sizeof dfn);    memset(sccno,0,sizeof sccno);}void addedge(int u,int v){    e[++nume].nxt = head[u];    e[nume].v = v;    head[u] = nume;}void dfs(int u){    dfn[u] = lown[u] = ++dfs_clk;    S.push(u);    for(int i = head[u]; i ; i = e[i].nxt){        int v = e[i].v;        if(!dfn[v]){            dfs(v);            lown[u] = min(lown[u],lown[v]);        }        else if(!sccno[v]){            lown[u] = min(lown[u],dfn[v]);        }    }    if(lown[u] == dfn[u]){        scc_no++;        while(true){            int x = S.top(); S.pop();            sccno[x] = scc_no;            if(x==u) break;        }    }}void find_scc(){    for(int i = 1; i <= n; i++){        if(!dfn[i])            dfs(i);    }}int dfs_dp(int x){    if(dp[x] != -1) return dp[x];    int ret = 0;    for(int i = 0; i < G[x].size(); i++){        int d = G[x][i];        ret = max(ret,dfs_dp(d)+num[x]);    }    if(G[x].size()==0) ret = num[x];    return dp[x] = ret;}void solve(){    for(int i = 1; i <= n; i++){        int d = sccno[i];        num[d]++;        for(int j = head[i]; j; j = e[j].nxt){            if(d!=sccno[e[j].v])                G[d].push_back(sccno[e[j].v]);        }    }    int ret = 0;    for(int i = 1; i <= scc_no; i++){        ret = max(ret,dfs_dp(i));    }    printf("%d\n",ret);}int main(){    int ncase;    cin >> ncase;    while(ncase--){        scanf("%d%d",&n,&m);        init();        while(m--){            int a,b;            scanf("%d%d",&a,&b);            addedge(a,b);        }        find_scc();        solve();    }    return 0;}