POJ 3190 Stall Reservations 【贪心 区间】

Stall Reservations

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4325 Accepted: 1547 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.

恩，题目大意就是说，给出每头牛的区间范围，一台机器在同一时间只能为一头牛服务，问最少需要多少台机器。不细心，赋错值卡了一天了。。。。

#include <iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define maxn 50050using namespace std;int r[maxn];struct node{    int s,e,p;};node cow[maxn];bool cmp(node a,node b){    if(a.s!=b.s)        return a.s<b.s;    return a.e<b.e;}struct Node{    int time,pos;    friend bool operator < (Node a,Node b)    {        if(a.time!=b.time)            return a.time > b.time;        else            return a.pos > b.pos;    }};priority_queue<Node>q;int main(){    int n;    while(~scanf("%d",&n))    {        while(!q.empty())            q.pop();        memset(r,0,sizeof(r));        for(int i=1;i<=n;++i)        {            scanf("%d%d",&cow[i].s,&cow[i].e);            cow[i].p=i;        }        sort(cow+1,cow+n+1,cmp);        int k=0;        Node now={cow[1].e,cow[1].p};        q.push(now);        r[cow[1].p]=++k;        for(int i=2;i<=n;++i)        {            now=q.top();            if(cow[i].s<=now.time)                r[cow[i].p]=++k;            else            {                r[cow[i].p]=r[now.pos];                q.pop();            }            now={cow[i].e,cow[i].p};            q.push(now);        }        printf("%d\n",k);        for(int i=1;i<=n;++i)            printf("%d\n",r[i]);    }    return 0;}