HDU 4635 Strongly connected (有向图的强连通分量)

Posted 2020-07-27 季末Despaire

Strongly connected

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Give a simple directed graph with N nodes and M edges. Please tell me the maximum number of the edges you can add that the graph is still a simple directed graph. Also, after you add these edges, this graph must NOT be strongly connected.
A simple directed graph is a directed graph having no multiple edges or graph loops.
A strongly connected digraph is a directed graph in which it is possible to reach any node starting from any other node by traversing edges in the direction(s) in which they point. 

 

 

Input

The first line of date is an integer T, which is the number of the text cases.
Then T cases follow, each case starts of two numbers N and M, 1<=N<=100000, 1<=M<=100000, representing the number of nodes and the number of edges, then M lines follow. Each line contains two integers x and y, means that there is a edge from x to y.

 

 

Output

For each case, you should output the maximum number of the edges you can add.
If the original graph is strongly connected, just output -1.

 

 

Sample Input

3

3 3

1 2

2 3

3 1

3 3

1 2

2 3

1 3

6 6

1 2

2 3

3 1

4 5

5 6

6 4

 

Sample Output

Case 1: -1

Case 2: 1

Case 3: 15

 

题解

问：最多加多少条边，使得原图不是强连通图

可以反向思考，假设原图已满边，即sum=n*(n-1),减去多少条边使其不是强连通图

直接加边sss=n*(n-1)-m;

用Tarjan求出强连通分量并缩点,将缩好的点分为两部分，一部分到另一部分没有边（即这个点的入度或出度为0），然后用sss减去这两部分能构成的变数。

假设第i个强连通分量有num[i]个点，则它与另一半的图所能连接的边数为num[i]*(n-num[i]),求这个的最小值，减去就行了。

代码

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;

const int MAXN = 100000+100;
/*
 * Tarjan算法
 * 复杂度O(N+M)
 */
struct Edge{
    int to,next;
}edge[MAXN];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];//Belong数组的值是1~scc
int Index,top;
int scc;//强连通分量的个数
bool Instack[MAXN];
int num[MAXN];//各个强连通分量包含点的个数，数组编号1~scc

void init()
{
    tot=0;
    memset(head,-1,sizeof(head));
}

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

void Tarjan(int u)
{
    int v;
    Low[u]=DFN[u]=++Index;
    Stack[top++]=u;
    Instack[u]=true;
    for(int i=head[u];i!=-1;i=edge[i].next){
        v=edge[i].to;
        if(!DFN[v]){
            Tarjan(v);
            if(Low[u]>Low[v]) Low[u]=Low[v];
        }
        else if(Instack[v]&&Low[u]>DFN[v])
            Low[u]=DFN[v];
    }
    if(Low[u]==DFN[u]){
        scc++;
        do{
            v=Stack[--top];
            Instack[v]=false;
            Belong[v]=scc;
            num[scc]++;
        }
        while(v!=u);
    }
}

void solve(int n)
{
    memset(DFN,0,sizeof(DFN));
    memset(Instack,false,sizeof(Instack));
    memset(num,0,sizeof(num));
    Index=top=scc=0;
    for(int i=1;i<=n;i++)
        if(!DFN[i])
           Tarjan(i);
}

int in[MAXN],out[MAXN];

int main()
{
    int T;
    int iCase=0;
    scanf("%d",&T);
    while(T--){
        iCase++;
        init();
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            addedge(u,v);
        }
        solve(n);
        if(scc==1){
            printf("Case %d: -1\n",iCase);
            continue;
        }
        for(int i=1;i<=scc;i++){
            in[i]=0;
            out[i]=0;
        }
        for(int u=1;u<=n;u++){
            for(int i=head[u];i!=-1;i=edge[i].next){
                int v=edge[i].to;
                if(Belong[u]==Belong[v]) continue;
                out[Belong[u]]++;
                in[Belong[v]]++;
            }
        }
        long long sss=(long long)n*(n-1)-m;
        long long ans=0;
        for(int i=1;i<=scc;i++){
            if(in[i]==0||out[i]==0)
                ans=max(ans,sss-(long long)num[i]*(n-num[i]));
        }
        printf("Case %d: %lld\n",iCase,ans);
    }
}