HDU 1827 - Summer Holiday (强连通 + 缩点)

思路

缩点之后有几个入度为0的就要通知几次，然后在那个强连通分量里找最便宜的打过去。

代码

#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <iomanip>
#include <cmath>
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const double eps = 1e-6;
const int MAXN = 1100 + 10;
const int MOD = 1000007;
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
int fee[MAXN];
struct SCC
{
    int v, pre[MAXN], lowlink[MAXN], sccno[MAXN], scc_cnt, dfs_clock, in[MAXN], out[MAXN];
    vector<int> G[MAXN], scc[MAXN];
    stack<int> st;
    void init(int v)
    {
        this->v = v;
        for (int i = 1; i <= v; i++) G[i].clear(), scc[i].clear();
        MS(sccno, 0); MS(pre, 0); MS(lowlink, 0); MS(in, 0);
        scc_cnt = dfs_clock = 0;
    }
    void DFS(int u)
    {
        pre[u] = lowlink[u] = ++dfs_clock;
        st.push(u);
        for (int i = 0; i < SZ(G[u]); 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 (pre[u] == lowlink[u])
        {
            ++scc_cnt;
            while (1)
            {
                int x = st.top(); st.pop();
                sccno[x] = scc_cnt;
                if (x == u) break;
            }
        }
    }
    void solve()
    {
        for (int i = 1; i <= v; i++)
        {
            scc[sccno[i]].PB(i);
            for (int j = 0; j < SZ(G[i]); j++)
            {
                int x = G[i][j];
                if (sccno[i] != sccno[x])
                {
                    out[sccno[i]]++;
                    in[sccno[x]]++;
                }
            }
        }
        int ans = 0, cnt = 0;
        for (int i = 1; i <= scc_cnt; i++)
        {
            int tmp = INF;
            if (in[i] == 0)
            {
                cnt++;
                for (int j = 0; j < SZ(scc[i]); j++)
                    tmp = min(tmp, fee[scc[i][j]]);
                ans += tmp;
            }
        }
        printf("%d %d\n", cnt, ans);
    }
    void find_scc()
    {
        for (int i = 1; i <= v; i++)
            if (!pre[i]) DFS(i);
    }
}scc;
int main()
{
    //ROP;
    int v, e, i, j;
    while (~scanf("%d%d", &v, &e))
    {
        scc.init(v);
        for (i = 1; i <= v; i++) scanf("%d", &fee[i]);
        for (i = 0; i < e; i++)
        {
            int a, b;
            scanf("%d%d", &a, &b);
            scc.G[a].PB(b);
        }
        scc.find_scc();
        scc.solve();
    }
    return 0;
}