uva 11324 The Largest Clique（图论-tarjan,动态规划）

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发布于 2022-07-06 17:04:54

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Problem B: The Largest Clique

Given a directed graph G, consider the following transformation. First, create a new graph T(G) to have the same vertex set as G. Create a directed edge between two vertices u and v in T(G) if and only if there is a path between u and v in G that follows the directed edges only in the forward direction. This graph T(G) is often called the transitive closure of G.

We define a clique in a directed graph as a set of vertices U such that for any two vertices u and v in U, there is a directed edge either from u to v or from v to u (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 graph G. It begins with a line of two integers n and m, where 0 ≤ n ≤ 1000 is the number of vertices of G and 0 ≤ m ≤ 50,000 is the number of directed edges of G. The vertices of G are numbered from 1 to n. The following m lines contain two distinct integers u and v between 1 and n which define a directed edge from u to v in G.

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

Sample input

Output for sample input

Zachary Friggstad

题目大意：

T组測试数据。给一张有向图G。求一个结点数最大的结点集，使得该结点中随意两个结点 u 和 v满足：要么 u 能够到达 v。要么 v 能够到达 u（u 和 v 相互可达也能够）。

解题思路：

”同一个强连通分量中的点要么都选，要么不选。把强连通分量收缩点后得到SCC图。让每一个SCC结点的权等于它的结点数，则题目转化为求SCC图上权最大的路径。因为SCC图是一个 DAG，能够用动态规划求解。 “

解题代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;

const int maxn=1100;
const int maxm=51000;

struct edge{
    int u,v,next;
    edge(int u0=0,int v0=0){
        u=u0;v=v0;
    }
}e[maxm];

int n,m,head[maxn],dfn[maxn],low[maxn],mark[maxn],w[maxn],color[maxn],dp[maxn],cnt,nc,index;
vector <int> vec;
vector <vector<int> > dfsmap;

void addedge(int u,int v){
    e[cnt]=edge(u,v);e[cnt].next=head[u];head[u]=cnt++;
}

void input(){
    cnt=nc=index=0;
    scanf("%d%d",&n,&m);
    vec.clear();
    for(int i=0;i<=n;i++){
        w[i]=dfn[i]=0;
        mark[i]=false;
        color[i]=dp[i]=head[i]=-1;
    }
    int u,v;
    while(m-- >0){
        scanf("%d%d",&u,&v);
        addedge(u,v);
    }
}

void tarjan(int s){
    dfn[s]=low[s]=++index;
    mark[s]=true;
    vec.push_back(s);
    for(int i=head[s];i!=-1;i=e[i].next){
        int d=e[i].v;
        if(!dfn[d]){
            tarjan(d);
            low[s]=min(low[d],low[s]);
        }else if(mark[d]){
            low[s]=min(low[s],dfn[d]);
        }
    }
    if(dfn[s]==low[s]){
        nc++;
        int d;
        do{
            d=vec.back();
            vec.pop_back();
            color[d]=nc;
            mark[d]=false;
            w[nc]++;
        }while(d!=s);
    }
}

int DP(int s){
    if(dp[s]!=-1) return dp[s];
    int ans=w[s];
    for(int i=0;i<dfsmap[s].size();i++){
        int d=dfsmap[s][i];
        if(DP(d)+w[s]>ans) ans=DP(d)+w[s];
    }
    return dp[s]=ans;
}

void solve(){
    for(int i=1;i<=n;i++){
        if(!dfn[i]) tarjan(i);
    }
    dfsmap.clear();
    dfsmap.resize(nc+1);
    for(int i=0;i<cnt;i++){
        int x=color[e[i].u],y=color[e[i].v];
        if(x!=y){
            dfsmap[x].push_back(y);
            //cout<<x<<"->"<<y<<endl;
        }
    }
    int ans=0;
    for(int i=1;i<=nc;i++){
        if(DP(i)>ans) ans=DP(i);
        //cout<<i<<" "<<ans<<endl;
    }
    printf("%d\n",ans);
}

int main(){
    int t;
    scanf("%d",&t);
    while(t-- >0){
        input();
        solve();
    }
    return 0;
}

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原始发表：2022年1月1，如有侵权请联系 cloudcommunity@tencent.com 删除

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