【强连通分量+缩点】 HDOJ 1827 Summer Holiday

先求出强连通分量，每个强连通分量都可以看成一个点，这个点的权值是这个连通分量重权值最小的，然后计算一下所有点的入度，入度为0的点的个数和权值和就是答案了。。

#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 1005
#define maxm 4005
#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], v[maxm], next[maxm];
int h[maxn], vv[maxm], nn[maxm];
int dfn[maxn], low[maxn];
int id[maxn], in[maxn], du[maxn];
int vis[maxn], n, m, top;
int w[maxn];
stack<int> s;

void read(void)
{
	int a, b, i, cnt = 0;
	for(i = 1; i <= n; i++) scanf("%d", &w[i]);
	while(m--) {
		scanf("%d%d", &a, &b);
		next[cnt] = H[a], v[cnt] = b, H[a] = cnt, cnt++;
	}
}
void init(void)
{
	top = 0;
	memset(H, -1, sizeof H);
	memset(h, -1, sizeof h);
	memset(du, 0, sizeof du);
	memset(in, 0 ,sizeof in);
	memset(vis, 0, sizeof vis);
	memset(dfn, 0, sizeof dfn);
}
void build(void)
{
	int i, cnt = 0;
	for(i = 1; i <= n; i++)
		for(int e = H[i]; ~e; e = next[e])
			nn[cnt] = h[id[i]], h[id[i]] = cnt, vv[cnt] = id[v[e]], cnt++;
}
void tarjan(int u)
{
	dfn[u] = low[u] = ++top;
	s.push(u), in[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(in[v[e]]) low[u] = min(low[u], dfn[v[e]]);
	}
	if(dfn[u] == low[u]) {
		int tmp = s.top(); s.pop(), in[tmp] = 0;
		while(tmp != u) {
			id[tmp] = u;
			w[u] = min(w[u], w[tmp]);
			tmp = s.top();
			s.pop();
			in[tmp] = 0;
		}
		id[tmp] = u;
	}
}
void work(void)
{
	int ans = 0, cost = 0;
	for(int i = 1; i <= n; i++)
		if(!dfn[i]) tarjan(i);
	build();
	for(int i = 1; i <= n; i++)
		for(int e = h[i]; ~e; e = nn[e])
			if(vv[e] != i)
				du[vv[e]]++;
	for(int i = 1; i <= n; i++)
		if(!vis[id[i]] && !du[id[i]])
			vis[id[i]] = 1, ans++, cost += w[id[i]];
	printf("%d %d\n", ans, cost);
}
int main(void)
{
	while(scanf("%d%d", &n, &m)!=EOF) {
		init();
		read();
		work();
	}
	return 0;
}