【强连通分量】 HDOJ 3639 Hawk-and-Chicken

先强连通缩点求出DAG图，求最大权值的时候，因为存在重边，也存在一个点从多条路径到达终点的情况。。所以要建反图，找入度为0的点DFS求出最大权。。这样用DFS就可以很好的解决上面两个问题。。。

#include <iostream>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <bitset>  
#include <cstdio>  
#include <algorithm>  
#include <cstring>  
#include <climits>  
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 5005
#define maxm 30005
#define eps 1e-10
#define mod 1000000009
#define INF 99999999  
#define lowbit(x) (x&(-x))  
//#define lson o<<1, L, mid  
//#define rson o<<1 | 1, mid+1, R  
typedef long long LL;
//typedef int LL;
using namespace std;

int H[maxn], NEXT[maxm], V[maxm];
int h[maxn], next[maxm], v[maxm];
int dfn[maxn], low[maxn], id[maxn];
int in[maxn], ins[maxn], p[maxn];
int w[maxn], res[maxn], vis[maxn];
int n, m, scc_cnt, top, mx, ans, ans_cnt, mmx;
vector<int> vec[maxn];
stack<int> s;

void init(void)
{
	top = scc_cnt = ans_cnt = mx = ans = 0;
	memset(h, -1, sizeof h);
	memset(H, -1, sizeof H);
	memset(w, 0, sizeof w);
	memset(in, 0, sizeof in);
	memset(ins, 0, sizeof ins);
	memset(dfn, 0, sizeof dfn);
	for(int i = 0; i < maxn; i++) vec[i].clear();
}
void read(void)
{
	int a, b, cnt = 0;
	scanf("%d%d", &n, &m);
	while(m--) {
		scanf("%d%d", &a, &b);
		a++, b++;
		NEXT[cnt] = H[a], H[a] = cnt, V[cnt] = b, cnt++;
	}
}
void tarjan(int u)
{
	dfn[u] = low[u] = ++top;
	s.push(u), ins[u] = 1;
	for(int e = H[u]; ~e; e = NEXT[e])
		if(!dfn[V[e]]) {
			tarjan(V[e]);
			low[u] = min(low[u], low[V[e]]);
		}
		else if(ins[V[e]]) low[u] = min(low[u], dfn[V[e]]);
	if(dfn[u] == low[u]) {
		int tmp = s.top(); s.pop(), ins[tmp] = 0;
		++scc_cnt;
		while(tmp != u) {
			id[tmp] = scc_cnt;
			tmp = s.top();
			s.pop();
			ins[tmp] = 0;
		}
		id[tmp] = scc_cnt;
	}
}
void narrow(void)
{
	int cnt = 0;
	for(int i = 1; i <= n; i++)
		for(int e = H[i]; ~e; e = NEXT[e])
			next[cnt] = h[id[V[e]]], h[id[V[e]]] = cnt, v[cnt] = id[i], cnt++;
}
void dfs(int u)
{
	if(vis[u]) return;
	vis[u] = 1;
	mx += w[u];
	if(mx > mmx) mmx = mx;
	for(int e = h[u]; ~e; e = next[e]) dfs(v[e]);
	//mx -= w[u];
}
void debug(void)
{
	printf("FD %d\n", ans_cnt);
	for(int i = 0; i < ans_cnt; i++)
		printf("AAA %d BBB\n", p[i]);
}
void work(int CASE)
{
	int i, num = 0;
	for(i = 1; i <= n; i++)
		if(!dfn[i]) tarjan(i);
	narrow();
	for(i = 1; i <= scc_cnt; i++)
		for(int e = h[i]; ~e; e = next[e])
			if(v[e] != i) in[v[e]]++;
	for(i = 1; i <= n; i++) w[id[i]]++;
	for(i = 1; i <= n; i++) vec[id[i]].push_back(i);
	for(i = 1; i <= scc_cnt; i++)
		if(!in[i]) {
			memset(vis, 0, sizeof vis);
			mmx = mx = 0;
			dfs(i);
			if(mmx == ans) p[ans_cnt++] = i;
			if(mmx > ans) ans_cnt = 0, p[ans_cnt++] = i, ans = mmx;
		}
	for(i = 0; i < ans_cnt; i++) {
		int d = vec[p[i]].size();
		for(int j = 0; j < d; j++)
			res[num++] = vec[p[i]][j];
	}
	sort(res, res+num);
	printf("Case %d: %d\n", CASE, --ans);
	if(num) printf("%d", --res[0]);
	for(i = 1; i < num; i++) printf(" %d", --res[i]);
	printf("\n");
}
int main(void)
{
	int _, __;
	while(scanf("%d", &_)!=EOF) {
		__ = 0;
		while(_--) {
			init();
			read();
			work(++__);
		}
	}
	return 0;
}