标签：chinese blog att std statement cas string user air
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.
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.
Print a single integer corresponding to the number of unordered pairs as asked in the problem..
Subtask 1: (30 points)
Subtask 2: (20 points)
Subtask 3 (50 points)
Input: 6 6 1 2 1 3 1 4 2 5 2 6 3 6 Output: 14
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)
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
https://s3.amazonaws.com/codechef_shared/download/translated/MAY15/mandarin/GRAPHCNT.pdf
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...
#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
CodeChef Counting on a directed graph
标签：chinese blog att std statement cas string user air
原文地址：http://www.cnblogs.com/neighthorn/p/6553009.html
