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

 template<class edge> struct Graph {
     vector<vector<edge> > adj;
     Graph(int n) {adj.clear(); adj.resize(n + );}
     Graph() {adj.clear(); }
     void resize(int n) {adj.resize(n + ); }
     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 << ];
     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 << ], minv[rt <<  | ]);
     }
     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 << ], minv[rt <<  | ]);
     }
     int get(int k, int l, int r, int rt) {
         if (l == r) {
             X = minv[rt];
             return l;
         }
         define_m;
         if (minv[rt <<  | ] <= 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 = , u, v; i < m; i++) {
             scanf("%d%d", &u, &v);
             G.add(u, v);
             H.add(v, u);
         }
         S.build(, n, );
         for (int i = n; i; i--) {
             int dot = S.get(k, , n, );
             k -= X;
             printf("%d%c", dot, i >= ? ' ' : '\n');
             S.update(dot, INF, , n, );
             for (int j = ; j < G[dot].size(); j++) {
                 S.update(G[dot][j], -, , n, );
             }
         }
     }
     return ;
 }