Codeforces Round #433 (Div. 2) C. Planning

题目链接：​​Planning​​

题目大意：有n架飞机，所有的飞机至少得在k+1分钟后才能飞行，第i架飞机每分钟延迟的费用是a[i]，每分钟只能起飞一架飞机，问最小延迟费用是多少

题目思路：按样例模拟，我们可以看到对于延迟花费最大的，我们应该让他在最接近的时候走，然后贪心做，用过的时间不能再用，所以要扔出去，用一个set维护就好了，思路比较乱，看代码一眼就能看懂了

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 3e5+10;

ll n,k,ans[maxn];
struct node{
    ll n1;
    ll n2;
    ll se;
}num[maxn];

bool cmp(node a,node b){
    if(a.n1 == b.n1) return a.n2 > b.n2;
    return a.n1 > b.n1;
}

int main(){
    while(~scanf("%I64d%I64d",&n,&k)){
        for(ll i = 0;i < n;i++){
            scanf("%I64d",&num[i].n1);
            num[i].n2 = i+1;
            num[i].se = i;
        }
        sort(num,num+n,cmp);
        set<ll>s;
        set<ll>::iterator it;
        ll sum = 0;
        for(ll i = 1;i <= n;i++) s.insert(i);
        for(ll i = 0;i < n;i++){
            ll p = *s.lower_bound(num[i].n2-k);
            sum = sum + (p+k-num[i].n2)*num[i].n1;
            ans[num[i].se] = p+k;
            s.erase(p);
        }
        printf("%I64d\n",sum);
        for(ll i = 0;i < n;i++) printf("%I64d ",ans[i]);
        puts("");
    }
    return 0;
}