[hdu5195]线段树

题意：给一个DAG，最多可以删去k条边，求字典序最大的拓扑序列。思路：贪心选取当前可选的最大编号即可，同时用线段树维护下。一个节点可以被选，当且仅当没有指向它的边。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define Rep(a, b) for(int a = 0; a < b; a++)
#define lowbit(x) ((x) & (-(x)))
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}

typedef double db;
typedef long long LL;

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, -1, 0, 1};
const int maxn = 1e5 + 7;
const int maxm = 1e5 + 7;
const int MD = 1e9 +7;
const int INF = 1e9 + 7;

template<class edge> struct Graph {
    vector<vector<edge> > adj;
    Graph(int n) {adj.clear(); adj.resize(n + 5);}
    Graph() {adj.clear(); }
    void resize(int n) {adj.resize(n + 5); }
    void add(int s, edge e){adj[s].push_back(e);}
    void del(int s, edge e) {adj[s].erase(find(iter(adj[s]), e)); }
    vector<edge>& operator [](int t) {return adj[t];}
};
Graph<int> G, H;
int X;
struct Seg {
    int minv[maxn << 2];
    void build(int l, int r, int rt) {
        if (l == r) {
            minv[rt] = H[l].size();
            return ;
        }
        define_m;
        build(lson);
        build(rson);
        minv[rt] = min(minv[rt << 1], minv[rt << 1 | 1]);
    }
    void update(int p, int x, int l, int r, int rt) {
        if (l == r) {
            minv[rt] += x;
            return ;
        }
        define_m;
        if (p <= m) update(p, x, lson);
        else update(p, x, rson);
        minv[rt] = min(minv[rt << 1], minv[rt << 1 | 1]);
    }
    int get(int k, int l, int r, int rt) {
        if (l == r) {
            X = minv[rt];
            return l;
        }
        define_m;
        if (minv[rt << 1 | 1] <= k) return get(k, rson);
        return get(k, lson);
    }
};
Seg S;
int main() {
    //freopen("in.txt", "r", stdin);
    int n, m, k;
    while (cin >> n >> m >> k) {
        G.adj.clear();
        G.resize(n);
        H.adj.clear();
        H.resize(n);
        for (int i = 0, u, v; i < m; i++) {
            scanf("%d%d", &u, &v);
            G.add(u, v);
            H.add(v, u);
        }
        S.build(1, n, 1);
        for (int i = n; i; i--) {
            int dot = S.get(k, 1, n, 1);
            k -= X;
            printf("%d%c", dot, i >= 2? ' ' : '
');
            S.update(dot, INF, 1, n, 1);
            for (int j = 0; j < G[dot].size(); j++) {
                S.update(G[dot][j], -1, 1, n, 1);
            }
        }
    }
    return 0;
}