HDU 4635 Strongly connected （2013多校4 1004 有向图的强连通分量）

Strongly connected

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 53    Accepted Submission(s): 15

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

 

 

Source

2013 Multi-University Training Contest 4

 

 

Recommend

zhuyuanchen520

 

 Tarjan 缩点。

 

/*
 *  Author:kuangbin
 *  1004.cpp
 */

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>
#include <map>
#include <vector>
#include <queue>
#include <set>
#include <string>
#include <math.h>
using namespace std;
/*
 * Tarjan算法
 * 复杂度O(N+M)
 */
const int MAXN = 100010;//点数
const int MAXM = 100010;//边数
struct Edge
{
    int to,next;
}edge[MAXM];
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
//num数组不一定需要，结合实际情况

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 = scc = top = 0;
    for(int i = 1;i <= N;i++)
        if(!DFN[i])
            Tarjan(i);
}
void init()
{
    tot = 0;
    memset(head,-1,sizeof(head));
}
int in[MAXN],out[MAXN];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    scanf("%d",&T);
    int iCase = 0;
    int n,m;
    int u,v;
    while(T--)
    {
        iCase++;
        init();
        scanf("%d%d",&n,&m);
        for(int i = 0;i < m;i++)
        {
            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: %d\n",iCase,ans);
    }
    return 0;
}