强连通缩点— HDU1827

　　强连通缩点以后最终形成的是一棵树

　　我们可以根据树的性质来看缩点以后的强连通分量图，就很好理解了

/*  gyt
       Live up to every day            */
#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<cstring>
#include<queue>
#include<set>
#include<string>
#include<map>
#include <time.h>
#define PI acos(-1)
using namespace std;
typedef long long ll;
typedef double db;
const int maxn = 4000+10;
const ll maxm = 1e7;
const int modd = 10000007;
const int INF = 1<<30;
const db eps = 1e-9;
struct Edge{
    int u, v, next;
}e[maxn*4];
int n, m, cnt,scnt, tot;
stack<int>sta;
int dfn[maxn], low[maxn], vis[maxn], head[maxn];
int du[maxn], color[maxn];
int a[maxn], in[maxn];

void init() {
    memset(head, -1, sizeof(head));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(vis, 0, sizeof(vis));
    memset(du, 0, sizeof(du));
    memset(color, 0, sizeof(color));
    for (int i=0; i<maxn; i++)  in[i]=INF;
    while(!sta.empty())  sta.pop();
    cnt=0;
    scnt=0;  tot=0;
}
void add(int u, int v) {
    e[cnt].v=v, e[cnt].next=head[u];
    head[u]=cnt++;
}
void Tarjan(int s) {
    int minn, t;
    dfn[s]=low[s]=++tot;
    vis[s]=2;
    sta.push(s);
    for (int i=head[s]; ~i; i=e[i].next) {
        int t=e[i].v;
        if (!dfn[t]) {
            Tarjan(t);
            low[s]=min(low[s], low[t]);
        } else {
            if (vis[t]==2) {
                low[s]=min(low[s], dfn[t]);
            }
        }
    }
    if (low[s]==dfn[s]) {
        scnt++;
        while(!sta.empty()) {
            int t=sta.top();
            sta.pop();
            vis[t]=1;
            color[t]=scnt;
            if (t==s)  break;
        }
    }
}
void solve() {
    while(scanf("%d%d", &n, &m)!=EOF) {
        for (int i=1; i<=n; i++) {
            scanf("%d", a+i);
        }
        init();
        for (int i=0; i<m; i++) {
            int a, b;  scanf("%d%d", &a, &b);
            add(a, b);
        }
        for (int i=1; i<=n; i++) {
            if (!dfn[i])  Tarjan(i);
        }
         for (int u=1; u<=n; u++) {
           for (int i=head[u]; ~i; i=e[i].next) {
                int v=e[i].v;
                if (color[u]!=color[v]) {
                    du[color[v]]++;
                }
            }
        }
        int ans=0, ansnum=0;
        for (int i=1; i<=scnt; i++) {
            if (!du[i]) {
                ans++;
                int num=INF;
                for (int j=1; j<=n; j++) {
                    if (color[j]==i) {
                        in[i]=min(a[j], in[i]);
                    }
                }
                ansnum+=in[i];
            }
        }
        printf("%d %d
", ans, ansnum);
    }
}
int main() {
    int t = 1;
    //freopen("in.txt", "r", stdin);
    //scanf("%d", &t);
    while(t--)
        solve();
    return 0;
}