# [HAOI2006]受欢迎的牛 tarjan缩点 BZOJ1051

## 输入输出格式

 第一行：两个用空格分开的整数：N和M

 第二行到第M + 1行：每行两个用空格分开的整数：A和B，表示A喜欢B

 第一行：单独一个整数，表示明星奶牛的数量

## 输入输出样例

3 3
1 2
2 1
2 3

1

## 说明

【数据范围】

10%的数据N<=20, M<=50

30%的数据N<=1000,M<=20000

70%的数据N<=5000,M<=50000

100%的数据N<=10000,M<=50000

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 400005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}

ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/

int n, m;
int idx;
int col[maxn], dp[maxn], sum[maxn];
int sk[maxn], top;
int dfn[maxn], low[maxn];
int tot;
int vis[maxn];
int val[maxn];
int num[maxn];

struct node {
int u, v, nxt;
}edge[maxn];

int cnt;
void addedge(int x, int y) {
}

void tarjan(int x) {
sk[++top] = x; vis[x] = 1;
low[x] = dfn[x] = ++idx;
for (int i = head[x]; i; i = edge[i].nxt) {
int v = edge[i].v;
if (!dfn[v]) {
tarjan(v);
low[x] = min(low[x], low[v]);
}
else if (vis[v]) {
low[x] = min(low[x], dfn[v]);
}
}
if (dfn[x] == low[x]) {
tot++;
while (sk[top + 1] != x) {
col[sk[top]] = tot; sum[tot] += val[sk[top]]; vis[sk[top--]] = 0;
num[tot]++;
}
}
}

int deg[maxn];

int main()
{
//ios::sync_with_stdio(0);
rdint(n); rdint(m);
for (int i = 1; i <= m; i++) {
int u, v; rdint(u); rdint(v); addedge(u, v);
}
for (int i = 1; i <= n; i++)if (!dfn[i])tarjan(i);
for (int i = 1; i <= n; i++) {
for (int j = head[i]; j; j = edge[j].nxt) {
int v = edge[j].v;
if (col[i] != col[v]) {
deg[col[i]]++;
}
}
}
int ans = 0;
int ct = 0;
for (int i = 1; i <= tot; i++) {
if (!deg[i]) {
ct++;
if (ct > 1) { cout << 0 << endl; return 0; }
ans = i;
}
}
cout << num[ans] << endl;
return 0;
}


posted @ 2018-12-23 23:26  NKDEWSM  阅读(57)  评论(0编辑  收藏