HDU 5638 Toposort 拓扑排序 优先队列

时间限制：1S / 空间限制：256MB

【在线测试提交传送门】

【问题描述】

    There is a directed acyclic graph with n vertices and m edges. You are allowed to delete exact k edges in such way that the lexicographically minimal topological sort of the graph is minimum possible.    给定一个有n个顶点，m条边的DAG（有向无环图），删除其中的K条边，使得这个图的拓扑排序的字典序最小。

【输入格式】

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:The first line contains three integers n, m and k (1≤n≤100000,0≤k≤m≤200000) -- the number of vertices, the number of edges and the number of edges to delete.For the next m lines, each line contains two integers ui and vi, which means there is a directed edge from ui to vi (1≤ui,vi≤n).You can assume the graph is always a dag. The sum of values of n in all test cases doesn't exceed 10^6. The sum of values of m in all test cases doesn't exceed 2×10^6.第一行，一个整数T，表示有T组测试数据，对于每组测试数据：第一行，包含3个整数n，m和k(1≤n≤100000,0≤k≤m≤200000) 。分别表示顶点数、边数和要删除的边数。接下来m行，每行包含两个整数ui和vi，表示一条从ui到vi的边(1≤ui,vi≤n)。输入数据保证是一个DGA，所有测试点中n的和不超过10^6，m的和不超过22×10^6。

【输出格式】

For each test case, output an integer S=(∑i=1ni⋅pi) mod (10^9+7), where p1,p2,...,pn is the lexicographically minimal topological sort of the graph.对于每组测试数据，输出一个整数S，等于i乘以pi的和，结果对10^9+7取余。其中pi表示字典序最小的拓扑序中的第i个元素。

【输入样例1】

34 2 01 21 34 5 12 13 14 12 32 44 4 21 22 33 41 4

【输出样例1】

302730

【题目来源】

HDU 5638

【解题思路】

删除的点一定是入度小于等于k，且编号最小的点，使用优先队列维护。

【参考代码】

#include<stdio.h>#include<iostream>#include<cstring>#include<vector>#include<queue>using namespace std;const int maxn = 2e5+7;const int mod = 1e9+7;vector<int> E[maxn],rE[maxn];int in[maxn];int inq[maxn];int vis[maxn];priority_queue<int,vector<int>,greater<int> >Q;void init(){    for(int i=0;i<maxn;i++)        E[i].clear(),rE[i].clear(),in[i]=0;    memset(inq,0,sizeof(inq));    memset(vis,0,sizeof(vis));}int main(){    int t;    scanf("%d",&t);    while(t--)    {        init();        int n,m,k;        scanf("%d%d%d",&n,&m,&k);        for(int i=0;i<m;i++)        {            int x,y;scanf("%d%d",&x,&y);            E[x].push_back(y);            rE[y].push_back(x);            in[y]++;        }        long long Ans = 0;        for(int i = 1 ; i <= n ; ++ i)        {            if(in[i]<=k)            {                Q.push( i );                inq[i] = 1;            }        }        int num = 1;        while(!Q.empty()){            int x = Q.top() ; Q.pop(); inq[x] = 0;            if(k >= in[x]){                vis[x] = 1 , k -= in[x];                Ans=(Ans+1ll*num*x)%mod;                num=num+1;                for(int i=0;i<E[x].size();i++){                    int v =E[x][i];                    if(vis[v]) continue;                    in[v]--;                    if(in[v] <= k&&!inq[v]){                        Q.push(v);                        inq[v] = 1;                    }                }            }        }        printf("%I64d\n",Ans);    }}