Hdoj 5195 DZY Loves Topological Sorting 【拓撲】+【線段樹】

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).

題意：給你n條邊，刪去不多於K條邊，使輸出的字典序最大！！
策略：我們每次都找小於等於當前K的較大的數輸出就好了，
需明白：1，每減去一個入度都是減去一條邊。
2：找到一個點之後，一定要將對應點的入度變爲最大值，以防後面還有可能被找到。
代碼：

#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都是代表的對應的點。
    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);//每次都是儘量選比較大的點
    return query(l, mid,pos<<1);
}

void topo(){
    for(int i = 0; i < n; ++ i){
        int temp = query(1, n, 1);
        k -= in[temp];//表示去掉幾個點
        ans.push_back(temp);
        update(temp, INF, 1, n, 1); //找到後就要更新
        in[temp] = INF; //一定要變爲正無窮
        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;
}