HDU 5638 拓扑排序+优先队列

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5638

题意：

给你一个DAG图，删除k条边，使得能个得到字典序尽可能小的拓扑排序

题解：

把拓扑排序的算法稍微改一下，如果某个顶点的入度小于k也把它加到优先队列里面去。

k减小后队列里面会有些点不满足<=k，直接踢出来就好了。

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<functional>
using namespace std;

typedef long long LL;
const int maxn = 101010;
const int maxm = maxn << 1;
const int mod = 1e9 + 7;
int n, m, k;

vector<int> G[maxn];
int ind[maxn], used[maxn],inq[maxn];

void init() {
    for (int i = 1; i <= n; i++) G[i].clear();
    memset(ind, 0, sizeof(ind));
    memset(inq, 0, sizeof(inq));
    memset(used, 0, sizeof(used));
}

int main() {
    int tc;
    scanf("%d", &tc);
    while (tc--) {
        scanf("%d%d%d", &n, &m, &k);
        init();
        for (int i = 0; i < m; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
            ind[v]++;
        }
        priority_queue<int,vector<int>,greater<int> > pq;
        for (int i = 1; i <= n; i++) {
            if (ind[i] <= k) pq.push(i),inq[i] = 1;
        }
        vector<int> ans;
        while (!pq.empty()) {
            while (ind[pq.top()]>k) {
                inq[pq.top()] = 0;
                pq.pop();
            }
            int x = pq.top(); pq.pop();
            inq[x] = 0;
            k -= ind[x];
            ans.push_back(x); used[x] = 1;
            for (int i = 0; i < G[x].size(); i++) {
                int v = G[x][i];
                ind[v]--;
                if (ind[v] <= k&&!inq[v]&&!used[v]) {
                    pq.push(v);
                    inq[v] = 1;
                }
            }
        }
        LL cnt = 0;
        for (int i = 0; i < ans.size(); i++) {
            cnt += (LL)(i + 1)*ans[i];
            cnt %= mod;
        }
        printf("%lld\n", cnt);
    }
    return 0;
}