POJ 3553 Task schedule【拓扑排序 + 优先队列 / 贪心】

Task schedule
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 515 Accepted: 309 Special Judge
Description

There are n preemptive jobs to be processed on a single machine. Each job j has a processing time pj and deadline dj. 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 Cj is easily determined.

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

max{Cj-dj, 0}.

Input

The first line contains a single integer n, 1 ≤ n ≤ 50000. Each of the next n lines contains two integers pj and dj, 0 ≤ pj ≤ 1000, 0 ≤ dj ≤ 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.

Output

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

Sample Input

2
4 1
4 0
1
1 2
Sample Output

1
2
Source

Northeastern Europe 2003, Western Subregion

【代码】：

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,n,x) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

using namespace std;

const int maxn = 1e6+5;
const int mod = 142857;

int n,m,num,u,v;
int p[maxn],d[maxn];
priority_queue<P> q;
vector<int> G[maxn];
int inDeg[maxn];

void topSort()
{
    int ok=0;
    while(!q.empty()) q.pop();
    for(int i=1; i<=n; i++) if(!inDeg[i]) q.push(P(d[i],i));
    while(!q.empty())
    {
        int now = q.top().second; q.pop();
        printf("%d
",now);
        for(int i=0;i<G[now].size();i++)
        {
            int nxt = G[now][i];
            if(--inDeg[nxt] == 0)
            {
                q.push(P(d[nxt],nxt));
            }
        }
    }
}

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d%d",p+i,d+i);
    scanf("%d",&m);
    memset(inDeg,0,sizeof(inDeg));
    for(int i=1;i<=n;i++) G[i].clear();
    for(int i=0;i<m;i++)
    {
        scanf("%d%d",&u,&v);
        G[u].push_back(v);
        inDeg[v]++;
    }
    topSort();
}