hdu 5360 贪心+优先队列

贪心的思路还是比较好想的，每次选择cur（已经邀请成功的人数）所在的区间中右端点最小的（因为右端点大的在后面可以邀请成功的几率大），然后很自然的想到可以用一个优先队列来维护这些区间，只要每次把左端点小于等于cur的区间放到优先队列中即可。

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

const int N = 100001;
bool visit[N];
int a[N];

struct Node 
{
    int r, id;
    Node(){}
    Node( int _r, int _id )
    {
        r = _r, id = _id;
    }
    bool operator < ( const Node & o ) const 
    {
        return r > o.r;
    }
};

vector<Node> v[N];
priority_queue<Node> q;

int main ()
{
    int t;
    scanf("%d", &t);
    while ( t-- )
    {
        int n;
        scanf("%d", &n);
        for ( int i = 0; i <= n; i++ )
        {
            v[i].clear();
        }
        for ( int i = 1; i <= n; i++ )
        {
            scanf("%d", a + i);
        }
        for ( int i = 1; i <= n; i++ )
        {
            int tmp;
            scanf("%d", &tmp);
            v[a[i]].push_back( Node( tmp, i ) );
        }
        memset( visit, 0, sizeof(visit) );
        int cur = 0, cnt = 1;
        while ( 1 )
        {
            for ( int i = 0; i < v[cur].size(); i++ )
            {
                q.push(v[cur][i]);
            }
            bool flag = false;
            while ( !q.empty() )
            {
                Node tt = q.top();
                q.pop();
                a[cnt++] = tt.id;
                visit[tt.id] = true;
                if ( tt.r >= cur )
                {
                    cur++;
                    flag = true;
                    break;
                }
            }
            if ( !flag )
            {
                break;
            }
        }
        for ( int i = 1; i <= n; i++ )
        {
            if ( !visit[i] )
            {
                a[cnt++] = i;
            }
        }
        printf("%d
", cur);
        for ( int i = 1; i < n; i++ )
        {
            printf("%d ", a[i]);
        }
        printf("%d
", a[n]);
    }
    return 0;
}