Hdoj 5195 DZY Loves Topological Sorting 【拓扑】+【线段树】

DZY Loves Topological Sorting

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 922 Accepted Submission(s): 269

Problem Description
A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge (u→v) from vertex u to vertex v, u comes before v in the ordering.
Now, DZY has a directed acyclic graph(DAG). You should find the lexicographically largest topological ordering after erasing at most k edges from the graph.

Input
The input consists several test cases. (TestCase≤5)
The first line, three integers n,m,k(1≤n,m≤105,0≤k≤m).
Each of the next m lines has two integers: u,v(u≠v,1≤u,v≤n), representing a direct edge(u→v).

Output
For each test case, output the lexicographically largest topological ordering.

Sample Input
5 5 2
1 2
4 5
2 4
3 4
2 3
3 2 0
1 2
1 3

Sample Output
5 3 1 2 4
1 3 2

Hint
Case 1.
Erase the edge (2->3),(4->5).
And the lexicographically largest topological ordering is (5,3,1,2,4).

题意：给你n条边，删去不多于K条边，使输出的字典序最大！！
策略：我们每次都找小于等于当前K的较大的数输出就好了，
需明白：1，每减去一个入度都是减去一条边。
2：找到一个点之后，一定要将对应点的入度变为最大值，以防后面还有可能被找到。
代码：

#include <cstdio>#include <cstring>#include <vector>#include <iostream>#include <algorithm>const int M = 1e5+5;const int INF = 0x3f3f3f3f;using namespace std;int c[M<<2], in[M];vector<int > m[M];vector<int > ans;int n, mm, k;void update(int p, int x, int l, int r, int pos){    if(l == r){        c[pos] = x; return ;    }    int mid = (l+r)>>1;    if(p <= mid) update(p, x, l, mid, pos<<1); //left和right都是代表的对应的点。    else update(p, x, mid+1, r, pos<<1|1);    c[pos] = min(c[pos<<1], c[pos<<1|1]);}int query(int l, int r, int pos){    if(l == r) return l;    int mid = (l+r)>>1;    if(c[pos<<1|1] <= k) return query(mid+1, r, pos<<1|1);//每次都是尽量选比较大的点    return query(l, mid,pos<<1);}void topo(){    for(int i = 0; i < n; ++ i){        int temp = query(1, n, 1);        k -= in[temp];//表示去掉几个点        ans.push_back(temp);        update(temp, INF, 1, n, 1); //找到后就要更新        in[temp] = INF; //一定要变为正无穷        for(int i = 0; i < m[temp].size(); ++ i){            int v = m[temp][i];            --in[v];             update(v, in[v], 1, n, 1);        }    }}int main(){    while(scanf("%d%d%d", &n, &mm, &k) == 3){        for(int i = 0; i <= n; ++ i){            m[i].clear(); in[i] = 0;        }        int u, v;        for(int i = 0; i < mm; ++ i){            scanf("%d%d", &u, &v);            ++in[v];            m[u].push_back(v);        }        for(int i = 1; i <= n; ++ i){            update(i, in[i], 1, n, 1);        }        ans.clear();        topo();        printf("%d", ans[0]);        for(int i = 1; i < n; ++ i)            printf(" %d", ans[i]);        printf("\n");    }    return 0;}