Luogu P43916 图的遍历

我们把“u点能够到达的最大点”转化为反向图中能到达u点的所有点里的最大值，可知缩点后满足无后效性。val[i]的初值设为连通分量i中的最大点。反向存图，tarjan缩点，拓扑序dp即可。

#include <iostream>
#include <cstdio>
#include <queue>
#define maxn 100100
using namespace std;
int n, m;
void read(int &x) {
    x = 0;
    char ch = getchar();
    while (!isdigit(ch))
        ch = getchar();
    while (isdigit(ch))
        x = x * 10 + (ch ^ 48),
        ch = getchar();
    return;
}
struct E {
    int to, nxt;
} edge[maxn], edge2[maxn];
int head[maxn], top, head2[maxn], top2;
inline void insert(int u, int v) {
    edge[++top] = (E) {v, head[u]};
    head[u] = top;
}
inline void insert2(int u, int v) {
    edge2[++top2] = (E) {v, head2[u]};
    head2[u] = top2;
}
int low[maxn], dfn[maxn], timer, c[maxn], cnt, sta[maxn], stp;
bool ins[maxn];
int val[maxn];
void tarjan(int u) {
    low[u] = dfn[u] = ++ timer;
    sta[++stp] = u, ins[u] = true;
    for (int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].to;
        if (!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if (ins[v])
            low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] == low[u]) {
        ++cnt;
        int x;
        do {
            x = sta[stp--];
            ins[x] = false;
            c[x] = cnt;
            val[cnt] = max(val[cnt], x);
        } while (x != u);
    }
}
int ind[maxn];
void build() {
    for (int u = 1; u <= n; ++u)
        for (int i = head[u]; i; i = edge[i].nxt) {
            int v = edge[i].to;
            if (c[u] != c[v])
                insert2(c[u], c[v]), ++ind[c[v]];
        }
}
void dp() {
    queue<int> que;
    for (int i = 1; i <= cnt; ++i)
        if (!ind[i]) que.push(i);
    while (!que.empty()) {
        int u = que.front(); que.pop();
        for (int i = head2[u]; i; i = edge2[i].nxt) {
            int v = edge2[i].to;
            val[v] = max(val[v], val[u]);
            --ind[v];
            if (!ind[v])
                que.push(v);
        }
    }
    return;
}
int main() {
    read(n), read(m);
    int u, v;
    for (int i = 1; i <= m; ++i) {
        read(u), read(v);
        insert(v, u);
    }
    for (int i = 1; i <= n; ++i)
        if (!dfn[i])
            tarjan(i);
    build();
    dp();
    for (int i = 1; i <= n; ++i)
        printf("%d ", val[c[i]]);
    return 0;
}

至此luogu上真哥留下的缩点习题全部完成。晚上更新对最小树形图（朱刘算法）的理解。