HDU1827 电话通知

//可作为模板，静态邻接表
//HDU1827
/*
*State: HDU1827 125MS 336K 2082 B C++ 
*题目大意：
*        wiskey大牛要通过电话通知大家，如果你有别的队员的电话，就可以有你代劳
*        通知这些人，通知每个人的话费是不一样的，wiskey要花费最少达到通知所有
*        人的目的。打电话这种模型具有传递性，用<A, B>来表示A可以通知到B，即如果
*        <A, B>，且<B, C>，则<A, C>。
*解题思路：
*        强连通分量缩点，之后求入度为0的点的个数即为最小通知人数，求缩点后，每个
*        缩点的最小通知费用，再累加入度为0的缩点的费用即为最小费用。
*/

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int MAXN = 1005;
const int MAXE = 2005;
int dfn[MAXN], low[MAXN], inS[MAXN];
int step, scc, id[MAXN], myS[MAXN], top;
int weight[MAXN], head[MAXN], cntEdge;

typedef struct _node
{
    int v, next;
}N;
N edge[MAXE];
//缩点建有向无环图
int idw[MAXN];

void addEdge(int u, int v)
{
    edge[cntEdge].v = v;
    edge[cntEdge].next = head[u];
    head[u] = cntEdge++;
}

void init()
{
    cntEdge = top = 0;
    scc = 1;
    step = 0;
    for(int i = 0; i < MAXN; i++)
    {
        idw[i] = 0;
        id[i] = head[i] = -1;
        inS[i] = 0;
        dfn[i] = low[i] = -1;
    }
}

void tarjan_scc(int n)
{
    dfn[n] = low[n] = ++step;
    myS[top++] = n;
    inS[n] = 1;
    for(int f = head[n]; f != -1; f = edge[f].next)
    {
        int son = edge[f].v;
        if(dfn[son] == -1)
        {
            tarjan_scc(son);
            low[n] = min(low[n], low[son]);
        }
        else if(inS[son] == 1)
            low[n] = min(low[n], dfn[son]);
    }
    if(dfn[n] == low[n])
    {
        int tmp, Min = INT_MAX;
        do
        {
            tmp = myS[--top];
            id[tmp] = scc;
            if(weight[tmp] < Min)
                Min = weight[tmp];
            inS[tmp] = 0;
        }while(top != 0 && tmp != n);
        idw[scc] += Min;
        scc++;
    }
}

void deal_scc(int n, int &sol1, int &sol2)
{
    int inTree[MAXN] = {0};
    for(int i = 1; i <= n; i++)
    {
        for(int j = head[i]; j != -1; j = edge[j].next)
        {
            int u = i, v = edge[j].v;
            if(id[u] == id[v])
                continue;
            else
                inTree[id[v]]++;
        }
    }
    sol1 = sol2 = 0;
    for(int i = 1; i < scc; i++)
    {
        if(inTree[i] == 0)
        {
            sol1++;
            sol2 += idw[i];
        }
    }
    
}

int main(void)
{
#ifndef ONLINE_JUDGE
    freopen("inHDU1827.txt", "r", stdin);
#endif

    int n, m;
    while(scanf("%d %d", &n, &m) == 2)
    {
        init();
        int u, v;
        for(int i = 1; i <= n; i++)
            scanf("%d", &weight[i]);

        for(int i = 0; i < m; i++)
        {
            scanf("%d %d", &u, &v);
            addEdge(u, v);
        }
        for(int i = 1; i <= n; i++)
        {
            if(dfn[i] == -1)
                tarjan_scc(i);
        }
        int sol1, sol2;
        deal_scc(n, sol1, sol2);
        printf("%d %d\n", sol1, sol2);
    }
    return 0;
}