Country of Metropolia is holding Olympiad of Metrpolises soon. It mean that all jury members of the olympiad should meet together in Metropolis (the capital of the country) for the problem preparation process.
There are n + 1 cities consecutively numbered from 0 to n. City 0 is Metropolis that is the meeting point for all jury members. For each city from 1 to n there is exactly one jury member living there. Olympiad preparation is a long and demanding process that requires k days of work. For all of these k days each of the n jury members should be present in Metropolis to be able to work on problems.
You know the flight schedule in the country (jury members consider themselves important enough to only use flights for transportation). All flights in Metropolia are either going to Metropolis or out of Metropolis. There are no night flights in Metropolia, or in the other words, plane always takes off at the same day it arrives. On his arrival day and departure day jury member is not able to discuss the olympiad. All flights in Megapolia depart and arrive at the same day.
Gather everybody for k days in the capital is a hard objective, doing that while spending the minimum possible money is even harder. Nevertheless, your task is to arrange the cheapest way to bring all of the jury members to Metrpolis, so that they can work together for kdays and then send them back to their home cities. Cost of the arrangement is defined as a total cost of tickets for all used flights. It is allowed for jury member to stay in Metropolis for more than k days.
The first line of input contains three integers n, m and k (1 ≤ n ≤ 105, 0 ≤ m ≤ 105, 1 ≤ k ≤ 106).
The i-th of the following m lines contains the description of the i-th flight defined by four integers di, fi, ti and ci (1 ≤ di ≤ 106, 0 ≤ fi ≤ n, 0 ≤ ti ≤ n, 1 ≤ ci ≤ 106, exactly one of fi and ti equals zero), the day of departure (and arrival), the departure city, the arrival city and the ticket cost.
Output the only integer that is the minimum cost of gathering all jury members in city 0 for k days and then sending them back to their home cities.
If it is impossible to gather everybody in Metropolis for k days and then send them back to their home cities, output "-1" (without the quotes).
2 6 5 1 1 0 5000 3 2 0 5500 2 2 0 6000 15 0 2 9000 9 0 1 7000 8 0 2 6500
24500
2 4 5 1 2 0 5000 2 1 0 4500 2 1 0 3000 8 0 1 6000
-1
The optimal way to gather everybody in Metropolis in the first sample test is to use flights that take place on days 1, 2, 8 and 9. The only alternative option is to send jury member from second city back home on day 15, that would cost 2500 more.
In the second sample it is impossible to send jury member from city 2 back home from Metropolis.
题目大意:有n个人要去0城市参加会议，有m架航班从n个位置到0或者从0到n，现在要求n个人都到0城市后开k天的会回去，飞机当天不算在k天开会之间，要求算出最小花费
解题思路:这是一个贪心，但是D题一般是很难做的，这道题首先我们按照航班起飞时间对飞机进行排序，然后对于去0城市飞机统计第i天是否能够全部到达，然后倒着贪心回去的飞机，然后遍历飞机去0城市的第i天的最小花费，和回去的最小花费,之后求和的最小值
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long LL;
struct point
{
int d,s,e,c;
}a[1000005];
LL s[1000005],e[1000005],ans,sum;
int i,n,m,k,j,flag,x,MAX;
int check[1000005];
bool cmp(point x,point y){return x.d<y.d;}
int main()
{
cin>>n>>m>>k;
MAX =0;
for(i=1;i<=m;i++)
{
cin>>a[i].d>>a[i].s>>a[i].e>>a[i].c;
MAX=max(MAX,a[i].d);
}
sort(a+1,a+1+m,cmp);
flag=n;
for(i=1;i<=m;i++)
{
if(a[i].s)
{
if(check[a[i].s]==0)
{
check[a[i].s]=a[i].c;
flag--;
if(!flag)//说明截止到当前，他们都能飞到0城市
{
for(j=1;j<=n;j++)
s[a[i].d]+=check[j];
ans=s[a[i].d];
}
}
else if(a[i].c<check[a[i].s])
{
ans-=check[a[i].s]-a[i].c;
check[a[i].s]=a[i].c;
if(!flag)//若已经都能飞回去后，出现当前飞机更便宜，在之前的基础上sum,更新花费sum
{
s[a[i].d]=ans;
}
}
}
}
if(flag)
{
cout<<-1<<endl;
return 0;
}
memset(check,0,sizeof(check));
flag = n;
for(i=m;i>=1;i--)
{
if(a[i].e)
{
if(check[a[i].e]==0)
{
check[a[i].e]=a[i].c;
flag--;
if(flag==0)
{
for(j=1;j<=n;j++)
e[a[i].d]+=check[j];
ans=e[a[i].d];
}
}
else if(a[i].c<check[a[i].e])
{
ans-=check[a[i].e]-a[i].c;
check[a[i].e]=a[i].c;
if(flag==0)
{
e[a[i].d]=ans;
}
}
}
}
if(flag!=0)
{
cout<<-1<<endl;
return 0;
}
for(i=1;i<=MAX;i++)
{
if(s[i]==0)//如果当前i天无法判断是否已经到了，那么若是前一天i-1天可以判断出来，就s[i]=s[i-1]
s[i]=s[i-1];
else if(s[i-1])
s[i]=min(s[i],s[i-1]);
}
for(i=MAX;i>=1;i--)
{
if(e[i]==0)//同上
e[i]=e[i+1];
else if(e[i+1])
e[i]=min(e[i],e[i+1]);
}
sum=1e18;
for(i=1;i<=MAX-k-1;i++)
{
if(s[i]!=0&&e[i+k+1]!=0)
{
sum=min(sum,s[i]+e[i+k+1]);
}
}
if(sum==1e18)
cout<<-1<<endl;
else
cout<<sum<<endl;
return 0;
}