[HAOI2006]受欢迎的牛 tarjan缩点 BZOJ1051

题目背景

本题测试数据已修复。

题目描述

每头奶牛都梦想成为牛棚里的明星。被所有奶牛喜欢的奶牛就是一头明星奶牛。所有奶

牛都是自恋狂,每头奶牛总是喜欢自己的。奶牛之间的“喜欢”是可以传递的——如果A喜

欢B,B喜欢C,那么A也喜欢C。牛栏里共有N 头奶牛,给定一些奶牛之间的爱慕关系,请你

算出有多少头奶牛可以当明星。

输入输出格式

输入格式:

 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

输出格式:

 第一行:单独一个整数,表示明星奶牛的数量

输入输出样例

输入样例#1: 复制
3 3
1 2
2 1
2 3
输出样例#1: 复制
1

说明

只有 3 号奶牛可以做明星

【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

 

缩点之后看出度为0的点含有几个点即可;

注意如果满足出度=0的点个数>1,那么显然为0;

#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 head[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) {
	edge[++cnt].v = y; edge[cnt].nxt = head[x]; head[x] = cnt;
}

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(...)  评论(... 编辑 收藏