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;
}