hdu 5195 DZY Loves Topological Sorting （拓扑排序+线段树）

DZY Loves Topological Sorting

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1250    Accepted Submission(s): 403

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, ucomes 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).

 

思路：

用线段树维护区间最小值。

 

ps.之前没在查询操作加个pushup，一直跑不出答案，因为在查询过程中，sum的值是更新了的，所以要pushup下

实现代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid int m = (l + r) >> 1
const int M = 2e5+10;
const int inf = 0x3f3f3f3f;
int n,m,cnt,key;
vector<int>g[M];
int sum[M<<2],du[M];
void pushup(int rt){
    sum[rt] = min(sum[rt<<1],sum[rt<<1|1]);
}

void build(int l,int r,int rt){
    if(l == r){
        sum[rt] = du[l];
        return;
    }
    mid;
    build(lson);
    build(rson);
    pushup(rt);
}

void update(int p,int l,int r,int rt){
    if(l == r){
        sum[rt]--;
        return ;
    }
    mid;
    if(p <= m) update(p,lson);
    else  update(p,rson);
    pushup(rt);
}

void query(int l,int r,int rt){
    if(l == r){
        cnt -= sum[rt];
        key = l;
        sum[rt] = inf;
        return ;
    }
    mid;
    if(sum[rt<<1|1] <= cnt) query(rson);
    else query(lson);
    pushup(rt);
}

int main(){
    int u,v;
    while(cin>>n>>m>>cnt){
        for(int i = 1;i <= m;i ++){
            cin>>u>>v;
            g[u].push_back(v);
            du[v]++;  //入度
        }
        build(1,n,1);
        for(int i = 1;i <= n;i ++){
            query(1,n,1);
            if(i==1) cout<<key;
            else cout<<" "<<key;
            for(int j = 0;j < g[key].size();j++){
                int x = g[key][j];
                update(x,1,n,1);
            }
        }
        cout<<endl;
        memset(du,0,sizeof(du));
    }
    return 0;
}