HDU 1827 强连通 缩点 Summer Holiday

求出强连通分量，因为强连通中只要有一个人被通知到了，所有人都能被通知到。

缩点以后形成一个DAG，找出那些入度为0的点，累加上它们的权值就是答案。一个点的权值等于SCC中权值最小的那个点。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <stack>
using namespace std;

const int maxn = 1000 + 10;

int n, m;
int w[maxn], cost[maxn];
vector<int> G[maxn];

stack<int> S;
int pre[maxn], lowlink[maxn], sccno[maxn];
int dfs_clock, scc_cnt;

void dfs(int u)
{
    pre[u] = lowlink[u] = ++dfs_clock;
    S.push(u);
    for(int i = 0; i < G[u].size(); i++)
    {
        int v = G[u][i];
        if(!pre[v])
        {
            dfs(v);
            lowlink[u] = min(lowlink[u], lowlink[v]);
        }
        else if(!sccno[v]) lowlink[u] = min(lowlink[u], pre[v]);
    }
    if(lowlink[u] == pre[u])
    {
        scc_cnt++;
        for(;;)
        {
            int x = S.top(); S.pop();
            sccno[x] = scc_cnt;
            if(x == u) break;
        }
    }
}

void find_scc()
{
    dfs_clock = scc_cnt = 0;
    memset(sccno, 0, sizeof(sccno));
    memset(pre, 0, sizeof(pre));
    for(int i = 1; i <= n; i++) if(!pre[i]) dfs(i);
}

int indeg[maxn];

int main()
{
    while(scanf("%d%d", &n, &m) == 2)
    {
        for(int i = 1; i <= n; i++) G[i].clear();
        for(int i = 1; i <= n; i++) scanf("%d", w + i);
        while(m--)
        {
            int u, v; scanf("%d%d", &u, &v);
            G[u].push_back(v);
        }

        find_scc();

        memset(cost, 0x3f, sizeof(cost));
        for(int i = 1; i <= n; i++) cost[sccno[i]] = min(cost[sccno[i]], w[i]);

        memset(indeg, 0, sizeof(indeg));
        for(int i = 1; i <= n; i++)
        {
            for(int j = 0; j < G[i].size(); j++)
            {
                int u = sccno[i], v = sccno[G[i][j]];
                if(u == v) continue;
                indeg[v]++;
            }
        }

        int ans1 = 0, ans2 = 0;
        for(int i = 1; i <= scc_cnt; i++) if(!indeg[i]) { ans1++; ans2 += cost[i]; }
        printf("%d %d
", ans1, ans2);
    }

    return 0;
}