[HAOI2006]受歡迎的牛 tarjan縮點 BZOJ1051
阿新 • • 發佈:2018-12-24
題目背景
本題測試資料已修復。
題目描述
每頭奶牛都夢想成為牛棚裡的明星。被所有奶牛喜歡的奶牛就是一頭明星奶牛。所有奶
牛都是自戀狂,每頭奶牛總是喜歡自己的。奶牛之間的“喜歡”是可以傳遞的——如果A喜
歡B,B喜歡C,那麼A也喜歡C。牛欄裡共有N 頭奶牛,給定一些奶牛之間的愛慕關係,請你
算出有多少頭奶牛可以當明星。
輸入輸出格式
輸入格式: 第一行:兩個用空格分開的整數:N和M
第二行到第M + 1行:每行兩個用空格分開的整數:A和B,表示A喜歡B
輸出格式: 第一行:單獨一個整數,表示明星奶牛的數量
輸入輸出樣例
輸入樣例#1: 複製3 3 1 2 2 1 2 3輸出樣例#1: 複製
1
說明
只有 3 號奶牛可以做明星
【資料範圍】
10%的資料N<=20, M<=50
30%的資料N<=1000,M<=20000
70%的資料N<=5000,M<=50000
100%的資料N<=10000,M<=50000
縮點之後看出度為0的點含有幾個點即可;
注意如果滿足出度=0的點個數>1,那麼顯然為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; }