http://poj.org/problem?id=2942
各种逗。。。。
翻译白书上有；看了白书和网上的标程，学习了。。orz。
强连通分量就是先找出割点，然后用个栈在找出割点前维护子树，最后如果这个是割点那么子树就都是强连通分量，然后本题求的是奇圈，那么就进行黑白染色，判断是否为奇圈即可。将不是奇圈的所有强连通分量的点累计起来即可。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <stack>
#include <vector>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define printarr2(a, b, c) for1(_, 1, b) { for1(__, 1, c) cout << a[_][__]; cout << endl; }
#define printarr1(a, b) for1(_, 1, b) cout << a[_] << ‘\t‘; cout << endl
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
const int N=1005, M=1000005;
int ihead[N], n, m, cnt, LL[N], FF[N], mp[N][N], tot, s[M<<1], top, vis[N], ok[N], col[N];
struct ED { int from, to, next; } e[M<<1];
void add(int u, int v) {
e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; e[cnt].from=u;
e[++cnt].next=ihead[v]; ihead[v]=cnt; e[cnt].to=u; e[cnt].from=v;
}
bool ifind(int u) {
int v;
for(int i=ihead[u]; i; i=e[i].next) if(vis[v=e[i].to]) {
if(col[v]==-1) { col[v]=!col[u]; return ifind(v); }
else if(col[v]==col[u]) return true;
}
return false;
}
void color(int x) {
int y, u=e[x].from;
CC(vis, 0); CC(col, -1); col[u]=0;
do {
y=s[top--];
vis[e[y].from]=vis[e[y].to]=1;
} while(y!=x);
if(ifind(u)) for1(i, 1, n) if(vis[i]) ok[i]=1;
}
void tarjan(int u, int fa) {
FF[u]=LL[u]=++tot;
for(int i=ihead[u]; i; i=e[i].next) if(fa!=e[i].to) {
int v=e[i].to;
if(!FF[v]) {
s[++top]=i; //入栈这里要注意。。不要在上边入栈。。
tarjan(v, u);
if(LL[v]>=FF[u]) color(i);
LL[u]=min(LL[u], LL[v]);
}
else if(LL[u]>FF[v]) s[++top]=i, LL[u]=FF[v]; //入栈这里要注意。。
}
}
int main() {
while(1) {
read(n); read(m); int ans=0;
if(n==0 && m==0) break;
CC(mp, 0); CC(ihead, 0); CC(LL, 0); CC(FF, 0); CC(ok, 0); top=cnt=tot=0;
rep(i, m) {
int u=getint(), v=getint();
mp[u][v]=mp[v][u]=1;
}
for1(i, 1, n) for1(j, i+1, n) if(!mp[i][j]) add(i, j);
for1(i, 1, n) if(!FF[i]) tarjan(i, -1);
for1(i, 1, n) if(!ok[i]) ++ans;
printf("%d\n", ans);
}
return 0;
}
Description
Being a knight is a very attractive career: searching for the Holy Grail, saving damsels in distress, and drinking with the other knights are fun things to do. Therefore, it is not very surprising that in recent years the kingdom of King Arthur has experienced an unprecedented increase in the number of knights. There are so many knights now, that it is very rare that every Knight of the Round Table can come at the same time to Camelot and sit around the round table; usually only a small group of the knights isthere, while the rest are busy doing heroic deeds around the country.
Knights can easily get over-excited during discussions-especially
after a couple of drinks. After some unfortunate accidents, King Arthur
asked the famous wizard Merlin to make sure that in the future no fights
break out between the knights. After studying the problem carefully,
Merlin realized that the fights can only be prevented if the knights are
seated according to the following two rules:
- The knights should
be seated such that two knights who hate each other should not be
neighbors at the table. (Merlin has a list that says who hates whom.)
The knights are sitting around a roundtable, thus every knight has
exactly two neighbors. - An odd number of knights should sit
around the table. This ensures that if the knights cannot agree on
something, then they can settle the issue by voting. (If the number of
knights is even, then itcan happen that ``yes" and ``no" have the same
number of votes, and the argument goes on.)
Merlin will let
the knights sit down only if these two rules are satisfied, otherwise he
cancels the meeting. (If only one knight shows up, then the meeting is
canceled as well, as one person cannot sit around a table.) Merlin
realized that this means that there can be knights who cannot be part of
any seating arrangements that respect these rules, and these knights
will never be able to sit at the Round Table (one such case is if a
knight hates every other knight, but there are many other possible
reasons). If a knight cannot sit at the Round Table, then he cannot be a
member of the Knights of the Round Table and must be expelled from the
order. These knights have to be transferred to a less-prestigious order,
such as the Knights of the Square Table, the Knights of the Octagonal
Table, or the Knights of the Banana-Shaped Table. To help Merlin, you
have to write a program that will determine the number of knights that
must be expelled.
Input
The
input contains several blocks of test cases. Each case begins with a
line containing two integers 1 ≤ n ≤ 1000 and 1 ≤ m ≤ 1000000 . The
number n is the number of knights. The next m lines describe which
knight hates which knight. Each of these m lines contains two integers
k1 and k2 , which means that knight number k1 and knight number k2 hate
each other (the numbers k1 and k2 are between 1 and n ).
The input is terminated by a block with n = m = 0 .
Output
For each test case you have to output a single integer on a separate line: the number of knights that have to be expelled.
Sample Input
5 5 1 4 1 5 2 5 3 4 4 5 0 0
Sample Output
2
Hint
Huge input file, ‘scanf‘ recommended to avoid TLE.
Source