POJ 3038 Flying Right

Flying Right

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1270		Accepted: 464

Description

Figuring that they cannot do worse than the humans have, Farmer John's cows have decided to start an airline. Being cows, they decide to cater to the heretofore-untapped market of cows as passengers. They plan to serve the cows who live along the western coast of Lake Michigan. Each morning, they will fly from the northern-most point of the coast southward towards Chicowgo, making many stops along the way. Each evening, they will fly back north to the northern-most point. 


They need your help to decide which passengers to carry each day. 
Each of N (1 <= N <= 10,000) farms numbered 1..N along the coast 
contains an airport (Farm 1 is northern-most; farm N is southern-most). 
On this day, K (1 <= K <= 50,000) groups of cows wish to 
travel.Each group of cows wants to fly from a particular farm to another
 particular farm.  The airline, if it wishes, is allowed to stop and 
pick up only part of a group. Cows that start a flight, however,must 
stay on the plane until they reach their destination.



Given the capacity C (1 <= C <= 100) of the airplane and the 
groups of cows that want to travel, determine the maximum number of cows
 that the airline can fly to their destination.

Input

* Line 1: Three space-separated integers: K, N, and C



* Lines 2..K+1: Each line contains three space-separated integers S,
 E, and M that specify a group of cows that wishes to travel. The M (1 
<= M <= C) cows are currently at farm S and want to travel to farm
 E (S != E).

Output

* Line 1: The maximum number of cows that can be flown to their destination. This is the sum of the number of cows flown to their destination on the flight southward in the morning plus the number of cows flown to their destination on the flight northward in the evening.

Sample Input

4 8 3
1 3 2
2 8 3
4 7 1
8 3 2

Sample Output

6

Hint

INPUT DETAILS: 
Four groups of cows, eight farms, and three seats on the plane.



OUTPUT DETAILS:

In the morning, the flight takes 2 cows from 1->3, 1 cow from 
2->8,and 1 cow from 4->7.  In the evening, the flight takes 2 cows
 from 8->3.

Source

USACO 2005 October Gold

//开始没看清题目、还以为是不能拆开的，原来一群牛可以选几头就可以了

//然后就贪心了，尽量不要选去远的牛、因为那样占据空间比较久
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <cmath>
#include <vector>
#define N 10004
#define Size 500004
using namespace std;
struct node
{
    int from,to,person;
    bool operator<(const node&a)const
    {
        return from<a.from;
    }
}to[Size],back[Size];
int off[N];
int cnt;
int C;
void del(node *go,int &nn)
{
    memset(off,0,sizeof(off));
    int i;
    int c=0,tp;
    int t1;
    priority_queue<int>qd;
    priority_queue<int,vector<int>,greater<int> >qx;
    for(i=0;i<nn;i++)
    {
        qd.push(go[i].to);
        qx.push(go[i].to);
        off[go[i].to]+=go[i].person;
        while(qx.top()<=go[i].from)
        {
            c-=off[qx.top()];
            off[qx.top()]=0;//尼玛（忍不住了），少了这句，从早上WA到现在
            qx.pop();
        }
        c+=go[i].person;
        cnt+=go[i].person;
        if(c>C)
        {
           tp=c-C;cnt-=tp;c=C;
           while(tp)
           {
               t1=qd.top();
               if(off[t1]>tp)
                {off[t1]-=tp;tp=0;}
               else
               {
                   tp-=off[t1];
                   off[t1]=0;
                   qd.pop();
               }
           }
        }
    }
}
int main()
{
    int S,E,M;
    int sn,en;
    int K,n;
    scanf("%d%d%d",&K,&n,&C);
     cnt=0;sn=en=0;
      while(K--)
       {
        scanf("%d%d%d",&S,&E,&M);
        if(S<E)
        {
           to[sn].from=S;
           to[sn].to=E;to[sn++].person=M;
        }
        else
        {
           back[en].from=E;
           back[en].to=S;back[en++].person=M;
        }
       }
       sort(to,to+sn);
       del(to,sn);
       sort(back,back+en);
       del(back,en);
       printf("%d\n",cnt);
    return 0;
}