【练习】HDU 3836 Equivalent Sets （强连通分量）

最新推荐文章于 2020-07-29 10:10:39 发布

原创最新推荐文章于 2020-07-29 10:10:39 发布 · 208 阅读

0 ·

CC 4.0 BY-SA版权

图论---强连通分量专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一个基于Tarjan算法实现的SCC（强连通分量）求解框架，通过缩点处理复杂图论问题。该框架能够高效地计算图中的强连通分量，并进一步统计各分量之间的连接特性，最终解决特定图论问题。

题意

同HDU 2767

题解

同HDU 2767

代码

#include<bits/stdc++.h>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int nmax = 1e6+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ull p = 67;
const ull MOD = 1610612741;
int N,M;
struct Tarjan{
    int head[nmax],low[nmax],dfn[nmax],sstack[nmax],color[nmax],top,sccnum,dfnnum,tot,n;
    int in[nmax],out[nmax];
    bool visit[nmax];
    struct edge{int to,nxt;}e[nmax<<1];
    void init(int n){
        memset(head,-1,sizeof head);
        memset(dfn,0,sizeof dfn);
        memset(color,0,sizeof color);
        memset(visit,0,sizeof visit);
        memset(in,0,sizeof in);
        memset(out,0,sizeof out);
        dfnnum = top = sccnum = tot = 0;
        this->n = n;
    }
    void add_edge(int u,int v){
        e[tot].to = v, e[tot].nxt = head[u]; head[u] = tot++;
    }
    void tarjan(int u){
        low[u] = dfn[u] = ++dfnnum;
        visit[u] = true;
        sstack[++top] = u;
        for(int i = head[u] ; i != -1; i = e[i].nxt){
            int v = e[i].to;
            if(!dfn[v]){
                tarjan(v);
                low[u] = min(low[u],low[v]);
            }else if(visit[v]) low[u] = min(low[u], dfn[v]);
        }
        if(dfn[u] == low[u]){
            visit[u] = false;
            color[u] = ++ sccnum;
            while(sstack[top] != u){
                color[sstack[top]] = sccnum;
                visit[sstack[top]] = false;
                top--;
            }
            top--;
        }
    }
    void dfs(int u){
        visit[u] = true;
        for(int i = head[u];i!=-1;i=e[i].nxt){
            int v = e[i].to;
            if(!visit[v]){
                if(color[v] != color[u]) in[color[v]]++ ,out[color[u]]++;
                dfs(v);
            }else if(visit[v]){
                if(color[v] != color[u]) in[color[v]]++ ,out[color[u]]++;
            }
        }
    }
    void start(){for(int i = 1;i<=n;++i) if(!dfn[i])
        tarjan(i);}
    int get_ans(){
        if(sccnum == 1) return 0;
        memset(visit,0,sizeof visit);
        for(int i = 1;i<=n;++i) if(!visit[i]) dfs(i);
        int notin = 0, notout = 0;
        for(int i = 1;i<=sccnum;++i){
            if(!in[i]) notin++;
            if(!out[i]) notout++;
        }
        return max(notin,notout);
    }
}solver;
int main(){
    while(scanf("%d %d",&N,&M)!=EOF){
        solver.init(N);
        int u,v;
        for(int i = 1;i<=M;++i){
            scanf("%d %d",&u,&v);
            solver.add_edge(u,v);
        }
        solver.start();
        printf("%d\n",solver.get_ans());
    }
    return 0;
}