HDU - 1827 Summer Holiday(强连通分量＋贪心)

题目大意：To see a World in a Grain of Sand
And a Heaven in a Wild Flower,
Hold Infinity in the palm of your hand
And Eternity in an hour.
―― William Blake

听说lcy帮大家预定了新马泰7日游，Wiskey真是高兴的夜不能寐啊，他想着得快点把这消息告诉大家，虽然他手上有所有人的联系方式，但是一个一个联系过去实在太耗时间和电话费了。他知道其他人也有一些别人的联系方式，这样他可以通知其他人，再让其他人帮忙通知一下别人。你能帮Wiskey计算出至少要通知多少人，至少得花多少电话费就能让所有人都被通知到吗？

解题思路：求出所有的强连通分量，接着缩点，然后找出所有强连通之间的关系。
找出所有入度为零的强连通分量（这些是必须要找一个人通知的），并找出这些强连通分量中价值最小的点

#include <cstdio>
#include <cstring>

#define max(a,b) ((a) > (b) ? (a) : (b))
#define min(a,b) ((a) < (b) ? (a) : (b))
#define N 1010
#define M 2010

struct Edge{
    int to, from, next, id;
}E[M];

int head[N], cost[N], pre[N], lowlink[N], sccno[N], stack[N], in[N], out[N], money[N];
int n, m, tot, scc_cnt, top, dfs_clock;;
bool vis[N];

void AddEdge(int from, int to) {
    E[tot].from = from;
    E[tot].to = to;
    E[tot].next = head[from];
    E[tot].id = tot;
    head[from] = tot++; 
}

void init() {
    memset(head, -1, sizeof(head));
    tot = 0;

    for (int i = 1; i <= n; i++)
        scanf("%d", &cost[i]);

    int u, v;
    for (int i = 0; i < m; i++) {
        scanf("%d%d", &u, &v);
        AddEdge(u, v);
    }
}

void dfs(int u) {
    pre[u] = lowlink[u] = ++dfs_clock;
    stack[++top] = u;

    for (int i = head[u]; i != -1; i = E[i].next) {
        int v = E[i].to;
        if (!pre[v]) {
            dfs(v);
            lowlink[u] = min(lowlink[u], lowlink[v]);
        }
        else if (!sccno[v]) {
            lowlink[u] = min(lowlink[u], pre[v]);
        }
    }

    int x;
    if (pre[u] == lowlink[u]) {
        scc_cnt++;

        while (1) {
            x = stack[top--];
            sccno[x] = scc_cnt;
            if (x == u)
                break;
        }
    }
}

void solve() {
    memset(pre, 0, sizeof(pre));
    memset(sccno, 0, sizeof(sccno));
    dfs_clock = top = scc_cnt = 0;

    for (int i = 1; i <= n; i++)
        if (!pre[i])
            dfs(i);

    for (int i = 1; i <= scc_cnt; i++)
        in[i] = 1;

    memset(money, 0x3f, sizeof(money));

    for (int i = 1; i <= n; i++) 
        money[sccno[i]] = min(money[sccno[i]], cost[i]);

    int u, v;
    for (int i = 0; i < tot; i++) {
        u = E[i].from, v = E[i].to;
        if (sccno[u] != sccno[v])
            in[sccno[v]] = 0;
    }

    int ans_man = 0, ans_cost = 0; 
    for (int i = 1; i <= scc_cnt; i++)
        if (in[i]) {
            ans_man++;
            ans_cost += money[i];
        }
    printf("%d %d\n", ans_man, ans_cost);
}

int main() {
    int cas = 1;
    while (scanf("%d%d", &n, &m) != EOF) {
        init();
        solve();
    }
    return 0;
}