HDU4635 Strongly connected

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

Source

2013 Multi-University Training Contest 4

Recommend

zhuyuanchen520

题意：给定N个点，M条边。问最多再添几条有向边使他恰好不为强连通图。

思路：假设有结果图。则必然是分成两块区域，且其中一区域的点有另一区域的全部路径，而另一区域无回路。

分别设为A,B。A中含有a条边，B中含有b条边。a+b = N。则结果图的总边数为all = a*(a-1)+ b*(b-1) + a*b ；

化简后 all = N*N- N - a*b 。 已知a+b = N 。 求all 最大值 ， 则 a ,b 相差越大， a*b越小 .(证a= (N-x)/2 , b = (N+x)/2 ) 。

A,B前提必是强连通分量，且入度或出度为0。

回到题中， 对给定的连通图，用tarjan算法求强连通个数（并记录每个强连通分量中点的个数），若原图为强连通图，输出-1;

否则对原图缩点，求出入度，出度。再求入度或出度为0的点中缩点前点集元素最少个数。

ans = all - M;

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
#define maxn 100000+10
using namespace std;
int dfn[maxn] , low[maxn] ;
int belong[maxn] ;
int indegree[maxn] , outdegree[maxn] ;
int times , bcnt ;
bool instack[maxn] ;
int snum[maxn] ;
int n , m ,minn;
vector<int> mapp[maxn] ;
stack<int> S ;
void tarjan(int u) {
int temp ;
dfn[u] = low[u] = times++ ;
instack[u] = true ;
S.push(u) ;
for( int i = 0 ; i < mapp[u].size() ;++i){
temp = mapp[u][i] ;
if(!dfn[temp] ) {
tarjan(temp) ;
low[u] = low[u] > low[temp] ? low[temp] : low[u] ;
}
else if( instack[temp] ) {
low[u] = low[u] > dfn[temp] ? dfn[temp] : low[u] ;
}
}
if( low[u] == dfn[u] ) {
bcnt++;
do
{
temp = S.top() ;
S.pop() ;
instack[temp] = false ;
belong[temp] = bcnt ;
snum[bcnt]++;
}
while (temp != u)  ;
}
}
int solve() {
for( int i = 1; i <= n ; ++i ) {
if(!dfn[i]) {
tarjan(i) ;
}
}
if( bcnt == 1 ) {
return false ;
}
for( int i = 1; i <= n ; ++i ) {
for( int j = 0 ;j < mapp[i].size() ; ++j ) {
int  temp =  mapp[i][j] ;
if( belong[i] != belong[temp] ) {
indegree[belong[temp]]++ ;
outdegree[belong[i]]++ ;
}
}
}
for( int i = 1; i <= bcnt ; ++i ){
if( !indegree[i]  || !outdegree[i] )  {
minn = minn > snum[i] ? snum[i] : minn ;
}
}
return minn ;

}
void init() {
memset(dfn,0,sizeof(dfn)) ;
memset(indegree,0,sizeof(indegree)) ;
memset(outdegree,0,sizeof(outdegree)) ;
memset(instack, 0 , sizeof(instack)) ;
memset(belong,0,sizeof(belong)) ;
memset(snum,0,sizeof(snum)) ;
while(!S.empty()) {
S.pop() ;
}
for( int i = 0; i <= n ; ++i ) {
mapp[i].clear() ;
}
times = 0 ;
bcnt = 0 ;
}
int main () {
int t , ncase  = 1 ,num1, num2;
cin >> t ;
while(t--) {
scanf("%d%d",&n,&m) ;
init () ;
for(int i = 0 ; i < m ; ++i ) {
scanf("%d%d",&num1,&num2) ;
mapp[num1].push_back(num2) ;
}
minn = maxn ;
printf("Case %d: ",ncase++) ;
if(solve()) {
cout << n*n-n-(n-minn)*minn - m << endl ;
}
else cout << -1 << endl;
}
return 0;
}