POJ3553 Task schedule （拓扑排序+优先队列）经典

There are n preemptive jobs to be processed on a single machine. Each job j has a processing time p_j and deadline d_j. Preemptive constrains are specified by oriented graph without cycles. Arc (i,j) in this graph means that job i has to be processed before job j. A solution is specified by a sequence of the jobs. For any solution the completion time C_j is easily determined.

The objective is to find the optimal solution in order to minimize

max{C_j-d_j, 0}.

The first line contains a single integer n, 1 ≤ n ≤ 50000. Each of the next n lines contains two integers p_j and d_j, 0 ≤ p_j ≤ 1000, 0 ≤ d_j ≤ 1000000, separated by one or more spaces. Line n+2 contains an integer m (number of arcs), 0 ≤ m ≤ 10*n. Each of the next m lines contains two integers i and j, 1 ≤ i, j ≤ n.

Each of the n lines contains integer i (number of job in the optimal sequence).

#include<stdio.h>
#include<queue>
#include<vector>
using namespace std;
const int N = 50005 ;
struct node
{
    int id,d;
    friend bool operator<(node a,node b)
    {
        return b.d>a.d;
    }
};

vector<int>mapt[N];
int path[N],in[N],d[N];

void print(int n)
{
    while(n--)
    {
        printf("%d\n",path[n]);
    }
}
void tope(int n)
{
    int k=0;
    priority_queue<node>q;
    node pre,now;

    for(int i = 1; i <= n; i++)
    if(in[i]==0)
    {
        now.d = d[i]; now.id = i; q.push(now);
    }
    while(!q.empty())
    {
        pre = q.top(); q.pop();
        path[k++] = pre.id;
        int len = mapt[pre.id].size();
        for(int i = 0 ;i < len; i++)
        {
            now.id = mapt[pre.id][i];
            now.d = d[now.id];
            in[now.id]--;
            if(in[now.id]==0)
            {
                q.push(now);
            }
        }
    }
    print(k);
}
int main()
{
    int n,m,a,b;
    while(scanf("%d",&n)>0)
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%d%d",&a,&d[i]);
            in[i] = 0;
            mapt[i].clear();
        }
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d%d",&a,&b);
            mapt[b].push_back(a);
            in[a]++;
        }
        tope(n);
    }
}