CodeChef Counting on a directed graph

Counting on a directed graph Problem Code: GRAPHCNT

All submissions for this problem are available.

Read problems statements in Mandarin Chinese and Russian.

Given an directed graph with N nodes (numbered from 1 to N) and M edges, calculate the number of unordered pairs (X, Y) such there exist two paths, one from node 1 to node X, and another one from node 1 to node Y, such that they don't share any node except node 1.

Input

There is only one test case in one test file.

The first line of each test case contains two space separated integers N, M. Each of the next M lines contains two space separated integers u, v denoting a directed edge of graph G, from node u to node v. There are no multi-edges and self loops in the graph.

Output

Print a single integer corresponding to the number of unordered pairs as asked in the problem..

Constraints and Subtasks

  • 1 ≤ N ≤ 105
  • 0 ≤ M ≤ 5 * 105

Subtask 1: (30 points)

 

Subtask 2: (20 points)

  • N * M ≤ 50000000

 

Subtask 3 (50 points)

  • No additional constraints

Example

Input:
6 6
1 2
1 3
1 4
2 5
2 6
3 6

Output:
14

Explanation

There are 14 pairs of vertices as follows: 
(1,2) 
(1,3) 
(1,4) 
(1,5) 
(1,6) 
(2,3) 
(2,4) 
(2,6) 
(3,4) 
(3,5) 
(3,6) 
(4,5) 
(4,6) 
(5,6)

5★ztxz16

7★kevinsogo

http://discuss.codechef.com/problems/GRAPHCNT

dominator may15 medium-hard ztxz16

25-03-2015

2 secs

50000 Bytes

ADA, ASM, BASH, BF, C, C99 strict, CAML, CLOJ, CLPS, CPP 4.3.2, CPP 4.9.2, CPP14, CS2, D, ERL, FORT, FS, GO, HASK, ICK, ICON, JAVA, JS, LISP clisp, LISP sbcl, LUA, NEM, NICE, NODEJS, PAS fpc, PAS gpc, PERL, PERL6, PHP, PIKE, PRLG, PYPY, PYTH, PYTH 3.4, RUBY, SCALA, SCM chicken, SCM guile, SCM qobi, ST, TCL, TEXT, WSPC

 

题意：

https://s3.amazonaws.com/codechef_shared/download/translated/MAY15/mandarin/GRAPHCNT.pdf

分析：

建出支配树，然后统计1号节点的每个儿子内部点对数量，这就是不合法的点对数量，用总的点对数量减去不合法的就好了...

代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<stack>
using namespace std;

const int maxn=100000+5,maxm=500000+5;

int n,m,tot,f[maxn],fa[maxn],id[maxn],dfn[maxn],siz[maxn],node[maxn],semi[maxn],idom[maxn];
long long ans;

stack<int> dom[maxn];

struct M{
	
	int cnt,hd[maxn],to[maxm],nxt[maxm];
	
	inline void init(void){
		cnt=0;
		memset(hd,-1,sizeof(hd));
	}
	
	inline void add(int x,int y){
		to[cnt]=y;nxt[cnt]=hd[x];hd[x]=cnt++;
	}
	
}G,tr;

inline bool cmp(int x,int y){
	return dfn[semi[x]]<dfn[semi[y]];
}

inline int find(int x){
	if(f[x]==x)
		return x;
	int fx=find(f[x]);
	node[x]=min(node[f[x]],node[x],cmp);
	return f[x]=fx;
}

inline void dfs(int x){
	dfn[x]=++tot;id[tot]=x;
	for(int i=tr.hd[x];i!=-1;i=tr.nxt[i])
		if(!dfn[tr.to[i]])
			dfs(tr.to[i]),fa[tr.to[i]]=x;
}

inline void LT(void){
	dfs(1);dfn[0]=tot<<1;
	for(int i=tot,x;i>=1;i--){
		x=id[i];
		if(i!=1){
			for(int j=G.hd[x],v;j!=-1;j=G.nxt[j])
				if(dfn[G.to[j]]){
					v=G.to[j];
					if(dfn[v]<dfn[x]){
						if(dfn[v]<dfn[semi[x]])
							semi[x]=v;
					}
					else{
						find(v);
						if(dfn[semi[node[v]]]<dfn[semi[x]])
							semi[x]=semi[node[v]];
					}
				}
			dom[semi[x]].push(x);
		}
		while(dom[x].size()){
			int y=dom[x].top();dom[x].pop();find(y);
			if(semi[node[y]]!=x)
				idom[y]=node[y];
			else
				idom[y]=x;
		}
		for(int j=tr.hd[x];j!=-1;j=tr.nxt[j])
			if(fa[tr.to[j]]==x)
				f[tr.to[j]]=x;
	}
	for(int i=2,x;i<=tot;i++){
		x=id[i];
		if(semi[x]!=idom[x])
			idom[x]=idom[idom[x]];
	}
	idom[id[1]]=0;
}

signed main(void){
	tr.init();G.init();
	scanf("%d%d",&n,&m);
	for(int i=1,x,y;i<=m;i++)
		scanf("%d%d",&x,&y),tr.add(x,y),G.add(y,x);
	for(int i=1;i<=n;i++)
		f[i]=node[i]=i;
	LT();ans=1LL*tot*(tot-1);
	for(int i=tot;i>=2;i--){
		siz[id[i]]++; 
		if(idom[id[i]]!=1)
			siz[idom[id[i]]]+=siz[id[i]];
		else
			ans-=1LL*siz[id[i]]*(siz[id[i]]-1);
	}
	ans>>=1;
	printf("%lld\n",ans);
	return 0;
}

　　

By NeighThorn

 

posted @ 2017-03-15 10:50  NeighThorn  阅读(267)  评论(0)  编辑  收藏  举报