tarjan算法,反向建图,HUD:Hawk-and-Chicken

Problem - 3639 (hdu.edu.cn)

Problem Description
Kids in kindergarten enjoy playing a game called Hawk-and-Chicken. But there always exists a big problem: every kid in this game want to play the role of Hawk.
So the teacher came up with an idea: Vote. Every child have some nice handkerchiefs, and if he/she think someone is suitable for the role of Hawk, he/she gives a handkerchief to this kid, which means this kid who is given the handkerchief win the support. Note the support can be transmitted. Kids who get the most supports win in the vote and able to play the role of Hawk.(A note:if A can win
support from B(A != B) A can win only one support from B in any case the number of the supports transmitted from B to A are many. And A can't win the support from himself in any case.
If two or more kids own the same number of support from others, we treat all of them as winner.
Here's a sample: 3 kids A, B and C, A gives a handkerchief to B, B gives a handkerchief to C, so C wins 2 supports and he is choosen to be the Hawk.
 

Input
There are several test cases. First is a integer T(T <= 50), means the number of test cases.
Each test case start with two integer n, m in a line (2 <= n <= 5000, 0
 

Output
For each test case, the output should first contain one line with "Case x:", here x means the case number start from 1. Followed by one number which is the total supports the winner(s) get.
Then follow a line contain all the Hawks' number. The numbers must be listed in increasing order and separated by single spaces.
 

Sample Input

2 4 3 3 2 2 0 2 1 3 3 1 0 2 1 0 2
 

Sample Output

Case 1: 2 0 1 Case 2: 2 0 1 2
 

Author
Dragon
 

Source
2010 ACM-ICPC Multi-University Training Contest(19)——Host by HDU

解析 :

tarjan算法缩点,然后反响建图,dfs遍历统计

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
const int N = 5e3 + 5, M = 3e4 + 5;
int n, m;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], timestamp;
int id[N], siz[N], scc_cnt;
int stk[N], top;
bool instk[N];
int din[N];
vectorG[N];
int d[N];
bool v[N];

void add(int a, int b) {
	e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

void tarjan(int u) {
	dfn[u] = low[u] = ++timestamp;
	stk[++top] = u;
	instk[u] = 1;
	for (int i = h[u]; i != -1; i = ne[i]) {
		int j = e[i];
		if (!dfn[j]) {
			tarjan(j);
			low[u] = min(low[u], low[j]);
		}
		else if (instk[j]) {
			low[u] = min(low[u], dfn[j]);
		}
	}
	if (dfn[u] == low[u]) {
		scc_cnt++;
		int y;
		do {
			y = stk[top--];
			instk[y] = 0;
			id[y] = scc_cnt;
			siz[scc_cnt]++;
		} while (y != u);
	}
}
int sum = 0;
void dfs(int u) {
	for (int i = 0; i < G[u].size(); i++) {
		int j = G[u][i];
		if (v[j])continue;
		v[j] = 1;
		dfs(j);
		sum += siz[j];
	}
}

int main() {
	int T;
	cin >> T;
	int cnt = 0;
	while (T--) {
		cnt++;
		scanf("%d%d", &n, &m);
		memset(h, -1, sizeof h);
		memset(instk, 0, sizeof instk);
		memset(dfn, 0, sizeof dfn);
		memset(siz, 0, sizeof siz);
		timestamp = top = idx = scc_cnt = 0;
		memset(d, 0, sizeof d);
		memset(din, 0, sizeof din);
		for (int i = 1, a, b; i <= m; i++) {
			scanf("%d%d", &a, &b);
			add(a, b);
		}
		for (int i = 0; i < n; i++) {
			if (!dfn[i])
				tarjan(i);
		}
		for (int i = 0; i < n; i++) {
			for (int j = h[i]; j != -1; j = ne[j]) {
				int k = e[j];
				if (id[i] != id[k]) {
					din[id[i]]++;
					G[id[k]].push_back(id[i]);
				}
			}
		}
		int ans = 0;
		for (int i = 1; i <= scc_cnt; i++) {
			if (din[i] == 0) {
				dfs(i);
				d[i] = sum + siz[i] - 1;
				sum = 0;
				ans = max(ans, d[i]);
				for (int j = 0; j <= scc_cnt; j++) {
					v[j] = 0;
				}
			}
		}
		int flg = 0;
		printf("Case %d: %d\n", cnt, ans);
		for (int i = 0; i < n; i++) {
			if (d[id[i]] == ans) {
				if (flg)printf(" ");
				printf("%d", i);
				flg = 1;
			}
		}
		printf("\n");
		for (int i = 0; i <= n; i++)
			G[i].clear();
	}
	return 0;
}

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