POJ3190

Stall Reservations

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5822 Accepted: 2130 Special Judge

Description

Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviously, FJ must create a reservation system to determine which stall each cow can be assigned for her milking time. Of course, no cow will share such a private moment with other cows. 

Help FJ by determining:

• The minimum number of stalls required in the barn so that each cow can have her private milking period
• An assignment of cows to these stalls over time

Many answers are correct for each test dataset; a program will grade your answer.

Input

Line 1: A single integer, N 

Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.

Output

Line 1: The minimum number of stalls the barn must have. 

Lines 2..N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.

Sample Input

51 102 43 65 84 7

Sample Output

412324

Hint

Explanation of the sample: 

Here's a graphical schedule for this output: 

Time     1  2  3  4  5  6  7  8  9 10Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..Stall 3 .. .. c3>>>>>>>>> .. .. .. ..Stall 4 .. .. .. c5>>>>>>>>> .. .. ..

Other outputs using the same number of stalls are possible.

Source

USACO 2006 February Silver

抽象一下就是：有几个区间，不相交的区间可以放在一个隔间里，

要使隔间数最少，求每个区间所在的隔间号~

使用优先队列

#include<cstdio>#include<algorithm>#include<queue>#include<iostream>using namespace std;struct node{    int x,y,c;    friend bool operator<(node a,node b)    {        return a.y>b.y;//y从小到大    }}p[50005];bool cmp(node a,node b){    return a.x<b.x;}int t[50005];int main(){    int i,j,n;    scanf("%d",&n);    for(i=0;i<n;i++)    {        scanf("%d%d",&p[i].x,&p[i].y);        p[i].c=i;    }    int r=0;    sort(p,p+n,cmp);    priority_queue<node> q;    q.push(p[0]);    t[p[0].c]=++r;    for(i=1;i<n;i++)    {        node g=q.top();        if(g.y<p[i].x)        {            t[p[i].c]=t[g.c];            q.pop();            q.push(p[i]);        }        else        {            t[p[i].c]=++r;            q.push(p[i]);        }    }    printf("%d\n",r);    for(int i=0;i<n;i++)        printf("%d\n",t[i]);    return 0;}