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
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.
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.
If the original graph is strongly connected, just output -1.
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
Case 2: 1
Case 3: 15
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 = ;//点数
const int MAXM = ;//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 != -;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,,sizeof(DFN));
memset(Instack,false,sizeof(Instack));
memset(num,,sizeof(num));
Index = scc = top = ;
for(int i = ;i <= N;i++)
if(!DFN[i])
Tarjan(i);
}
void init()
{
tot = ;
memset(head,-,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 = ;
int n,m;
int u,v;
while(T--)
{
iCase++;
init();
scanf("%d%d",&n,&m);
for(int i = ;i < m;i++)
{
scanf("%d%d",&u,&v);
addedge(u,v);
}
solve(n);
if(scc == )
{
printf("Case %d: -1\n",iCase);
continue;
}
for(int i = ;i <= scc;i++)
{
in[i] = ;
out[i] = ;
}
for(int u = ;u <= n;u++)
for(int i = head[u];i != -;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-) - m;
long long ans = ;
for(int i = ;i <= scc;i++)
{
if(in[i]== || out[i] == )
ans = max(ans,sss - (long long)num[i]*(n-num[i]));
}
printf("Case %d: %d\n",iCase,ans);
}
return ;
}
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