洛谷 P1607 [USACO09FEB]庙会班车Fair Shuttle 解题报告

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原文链接：http://www.cnblogs.com/butterflydew/p/9547564.html

博客围绕P1607庙会班车问题展开，描述了班车只能跑一次、停靠N个地点，K组奶牛有乘车需求，班车容量为C的情况。给出输入输出格式，还提到该题有多种做法，可按左端点排序，用平衡树维护车上奶牛右端点，采用贪心算法求解。

P1607 [USACO09FEB]庙会班车Fair Shuttle

题目描述

Although Farmer John has no problems walking around the fair to collect prizes or see the shows, his cows are not in such good shape; a full day of walking around the fair leaves them exhausted. To help them enjoy the fair, FJ has arranged for a shuttle truck to take the cows from place to place in the fairgrounds.

FJ couldn't afford a really great shuttle, so the shuttle he rented traverses its route only once (!) and makes N (1 <= N <= 20,000) stops (conveniently numbered 1..N) along its path. A total of K (1 <= K <= 50,000) groups of cows conveniently numbered 1..K wish to use the shuttle, each of the M_i (1 <= M_i <= N) cows in group i wanting to ride from one stop S_i (1 <= S_i < E_i) to another stop E_i (S_i < E_i <= N) farther along the route.

The shuttle might not be able to pick up an entire group of cows (since it has limited capacity) but can pick up partial groups as appropriate.

Given the capacity C (1 <= C <= 100) of the shuttle truck and the descriptions of the groups of cows that want to visit various sites at the fair, determine the maximum number of cows that can ride the shuttle during the fair.

逛逛集市，兑兑奖品，看看节目对农夫约翰来说不算什么，可是他的奶牛们非常缺乏锻炼——如果要逛完一整天的集市，他们一定会筋疲力尽的。所以为了让奶牛们也能愉快地逛集市，约翰准备让奶牛们在集市上以车代步。但是，约翰木有钱，他租来的班车只能在集市上沿直线跑一次，而且只能停靠N(1 ≤N≤20000)个地点（所有地点都以1到N之间的一个数字来表示）。现在奶牛们分成K(1≤K≤50000)个小组，第i 组有Mi(1 ≤Mi≤N)头奶牛，他们希望从Si跑到Ti(1 ≤Si<Ti≤N)。

由于班车容量有限，可能载不下所有想乘车的奶牛们，此时也允许小里的一部分奶牛分开乘坐班车。约翰经过调查得知班车的容量是C(1≤C≤100)，请你帮助约翰计划一个尽可能满足更多奶牛愿望的方案。

输入输出格式

输入格式：

第一行：包括三个整数：K，N和C，彼此用空格隔开。

第二行到K+1行：在第i+1行，将会告诉你第i组奶牛的信息：Si，Ei和Mi，彼

此用空格隔开。

输出格式：

第一行：可以坐班车的奶牛的最大头数。

这题的做法还是比较多哒

一看是区间选不选之类的那不就是排个序然后想办法贪心贪心呗云云

按左端点排序

拿一颗平衡树维护在车上的奶牛的右端点

当有左端点进来时，权值小于这颗左端点的点下车

然后上车上到车满

然后比一比右端点，把右端点大的踢出去

Code:

#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <ctime>
const int N=5e4+10;
int kk,n,c;//k个区间,n个时间,c的容量
struct node
{
    int s,t,m;
    bool friend operator <(node n1,node n2)
    {
        return n1.s<n2.s;
    }
}gro[N];
int root,dat[N<<6],siz[N<<6],ch[N<<6][2],val[N<<6],tot;
#define ls ch[now][0]
#define rs ch[now][1]
void updata(int now)
{
    siz[now]=siz[ls]+siz[rs]+1;
}
void split(int now,int k,int &x,int &y)
{
    if(!now) {x=y=0;return;}
    if(dat[now]<=k)
    {
        x=now;
        split(rs,k,rs,y);
    }
    else
    {
        y=now;
        split(ls,k,x,ls);
    }
    updata(now);
}
int Merge(int x,int y)
{
    if(!x||!y) return x+y;
    if(val[x]>val[y])
    {
        ch[x][1]=Merge(ch[x][1],y);
        updata(x);
        return x;
    }
    else
    {
        ch[y][0]=Merge(x,ch[y][0]);
        updata(y);
        return y;
    }
}
int New(int k)
{
    dat[++tot]=k,val[tot]=rand(),siz[tot]=1;
    return tot;
}
void Insert(int k)
{
    int x,y;
    split(root,k,x,y);
    root=Merge(x,Merge(New(k),y));
}
void extrack(int k)
{
    int x,y,z;
    split(root,k,x,y);
    split(x,k-1,x,z);
    z=Merge(ch[z][0],ch[z][1]);
    root=Merge(x,Merge(z,y));
}
int mx()
{
    int now=root;
    while(rs) now=rs;
    return dat[now];
}
int mi()
{
    int now=root;
    while(ls) now=ls;
    return dat[now];
}
int main()
{
    srand(time(0));
    scanf("%d%d%d",&kk,&n,&c);
    for(int i=1;i<=kk;i++)
        scanf("%d%d%d",&gro[i].s,&gro[i].t,&gro[i].m);
    std::sort(gro+1,gro+1+kk);
    int ans=0;
    for(int i=1;i<=kk;i++)
    {
        int mmi,mmx;
        while(siz[root]&&(mmi=mi())<=gro[i].s) extrack(mmi),++ans;//下车
        while(gro[i].m&&siz[root]<c) Insert(gro[i].t),--gro[i].m;//上车
        while(gro[i].m&&(mmx=mx())>gro[i].t) extrack(mmx),Insert(gro[i].t),--gro[i].m;
        //踢人
    }
    printf("%d\n",ans+siz[root]);
    return 0;
}

2018.8.28

转载于:https://www.cnblogs.com/butterflydew/p/9547564.html