强联通总结

noip前两天开始学这玩意…………

强连通（模版 元问题byscy）

模板题，我感觉不难。

另外可以用来缩点，在开一个邻接表，不再在一个强联通分量的连边就好。

#include<bits/stdc++.h>
#define REP(i, a, b) for(register int i = (a); i < (b); i++)
#define _for(i, a, b) for(register int i = (a); i <= (b); i++)
using namespace std;

const int MAXN = 2e4 + 10;
const int MAXM = 2e5 + 10;
struct Edge{ int to, next; } e[MAXM];
int head[MAXN], tot, n, m;

int low[MAXN], dfn[MAXN], id;
int belong[MAXN], ins[MAXN], cnt;
int sta[MAXN], top;

void AddEdge(int from, int to)
{
    e[tot] = Edge{to, head[from]};
    head[from] = tot++;
}

void tarjan(int u)
{
    low[u] = dfn[u] = ++id;
    sta[++top] = u; ins[u] = 1;
    for(int i = head[u]; ~i; i = e[i].next)
    {
        int v = e[i].to;
        if(!dfn[v])
        {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(ins[v]) low[u] = min(low[u], low[v]);
    }
    
    if(dfn[u] == low[u])
    {
        ++cnt;
        while(1)
        {
            int v = sta[top--];
            ins[v] = 0;
            belong[v] = cnt;
            if(u == v) break;
        }
    }
}

int main()
{
    memset(head, -1, sizeof(head)); tot = 0;
    scanf("%d%d", &n, &m);
    _for(i, 1, m)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        AddEdge(u, v);
    }
    
    _for(i, 1, n)
        if(!dfn[i])
            tarjan(i);
    printf("%d
", cnt);
    
    return 0;
}

强连通入门2：添加最少边成为强连通图

看题的时候感觉是一个公式

但是想复杂了，1h推出一个错误的结论……

这样想，如果是强联通图，肯定每个点的出度和入度都至少为1

那么我们把出度和入度为0的点消灭就好了

一条边可以消灭一个入度为0的点和1个出度为0的点

设s1为入度为0的点，s2为出度为0的点

那么用min（s1, s2）条边连接入度为0的点和出度为0的点

然后用max(s1, s2) - min（s1, s2）条边消灭剩下的点

所以答案是max(s1, s2)

#include<bits/stdc++.h>
#define REP(i, a, b) for(register int i = (a); i < (b); i++)
#define _for(i, a, b) for(register int i = (a); i <= (b); i++)
using namespace std;

const int MAXN = 2e4 + 10;
const int MAXM = 2e5 + 10;
struct Edge{ int to, next; } e[MAXM];
int head[MAXN], tot;

int dfn[MAXN], low[MAXN], id;
int sta[MAXN], ins[MAXN], top;
int belong[MAXN], n, m, cnt;
int in[MAXN], out[MAXN];

void AddEdge(int from, int to)
{
    e[tot] = Edge{to, head[from]};
    head[from] = tot++;
}

void tarjan(int u)
{
    dfn[u] = low[u] = ++id;
    sta[++top] = u; ins[u] = 1;
    
    for(int i = head[u]; ~i; i = e[i].next)
    {
        int v = e[i].to;
        if(!dfn[v])
        {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(ins[v]) low[u] = min(low[u], low[v]);
    }
    
    if(dfn[u] == low[u])
    {
        cnt++;
        while(1)
        {
            int v = sta[top--];
            ins[v] = 0;
            belong[v] = cnt;
            if(u == v) break;
        }
    }
}

void init()
{
    tot = cnt = id = top = 0;
    memset(head, -1, sizeof(head)); 
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(ins, 0, sizeof(ins));
       memset(in, 0, sizeof(in));
    memset(out, 0, sizeof(out));
}

int main()
{
    int T;
    scanf("%d", &T);
    
    while(T--)
    {
        init();
        scanf("%d%d", &n, &m);
        _for(i, 1, m)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            AddEdge(u, v);
        }
        
        _for(i, 1, n)
            if(!dfn[i])
                tarjan(i);
        
        _for(u, 1, n)
            for(int i = head[u]; ~i; i = e[i].next)
            {
                int v = e[i].to;
                int uu = belong[u], vv = belong[v]; //新的点 
                if(uu == vv) continue;
                out[uu]++; in[vv]++;
            }
        
        if(cnt == 1) { puts("0"); continue; } //特判 
            
        int ans1 = 0, ans2 = 0;
        _for(i, 1, cnt) 
        {
            ans1 += (in[i] == 0);
            ans2 += (out[i] == 0);
        }
        printf("%d
", max(ans1, ans2));
    }
    
    return 0;
}

强连通入门4：The Bottom of a Graph

一开始以为看这个点出去的点和自己是不是连通分量就好了

然后一直WA

然后发现这个点能达到的点不只这个点出去的点……这样做是错的……

答案应该是缩点后出度为0的连通分量的所有点

#include<bits/stdc++.h>
#define REP(i, a, b) for(register int i = (a); i < (b); i++) 
#define _for(i, a, b) for(register int i = (a); i <= (b); i++) 
using namespace std;

const int MAXN = 5e3 + 10;
struct Edge{ int to, next; };
vector<Edge> e;
int head[MAXN], tot;
int n, m;

int sta[MAXN], ins[MAXN], belong[MAXN], top;
int dfn[MAXN], low[MAXN], out[MAXN], id, cnt;

void AddEdge(int from, int to)
{
    e.push_back(Edge{to, head[from]});
    head[from] = tot++;
}

void init()
{
    tot = id = top = cnt = 0;
    e.clear();
    memset(head, -1, sizeof(head));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(ins, 0, sizeof(ins));
    memset(out, 0, sizeof(out));
}

void tarjan(int u)
{
    dfn[u] = low[u] = ++id;
    sta[++top] = u; ins[u] = 1;
    
    for(int i = head[u]; ~i; i = e[i].next)
    {
        int v = e[i].to;
        if(!dfn[v])
        {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(ins[v]) low[u] = min(low[u], low[v]);
    }
    
    if(dfn[u] == low[u])
    {
        ++cnt;
        while(1)
        {
            int v = sta[top--];
            ins[v] = 0;
            belong[v] = cnt;
            if(u == v) break;
        }
    }
}


int main()
{
    while(~scanf("%d", &n) && n)
    {
        init();
        scanf("%d", &m);
        _for(i, 1, m)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            AddEdge(u, v);
        }
        
        _for(i, 1, n)
            if(!dfn[i])
                tarjan(i);
        
        _for(u, 1, n)
            for(int i = head[u]; ~i; i = e[i].next)
            {
                int v = e[i].to;
                int uu = belong[u], vv = belong[v];
                if(uu != vv) out[uu]++;
            }
        
        int first = 1;
        set<int> ans;
        _for(i, 1, cnt) if(!out[i]) ans.insert(i);
        _for(i, 1, n)
            if(ans.count(belong[i]))
            {
                if(first) first = 0; else putchar(' ');
                printf("%d", i);
            }
        puts("");
    }
    
    return 0;
}

还差两道题，待补……