Stall Reservations (poj 3190 贪心)

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Stall Reservations

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3394 Accepted: 1215 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

题意：给出每个奶牛挤奶的时间段，一个机器一次只能对一头奶牛工作，问至少需要多少台机器，并输出每头奶牛使用的机器编号。

思路：先按照每头牛的开始时间从小到大排序，维护一个优先队列，n头牛依次入队，时间结束早的先出队列，比较最早出来的牛的结束时间和当前准备入队的牛的开始时间，若前者小于后者，则当前的牛可以用前面空出来的机器，否者要添加一台新机器。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 50010#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b)  for(i = a; i <= b; i++)#define FRL(i,a,b)  for(i = a; i < b; i++)#define mem(t, v)   memset ((t) , v, sizeof(t))#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define DBG         pf("Hi\n")typedef long long ll;using namespace std;struct Cow{    int s,t;    int local;}cow[maxn];int cmp(Cow a,Cow b){    if (a.s==b.s) return a.t<b.t;    return a.s<b.s;}priority_queue<Cow>Q;int ans[maxn];int n;bool operator<(Cow a,Cow b){    return a.t>b.t;}int main(){    int i,j;    Cow st;    while (~sf(n))    {        FRL(i,0,n)        {            sff(cow[i].s,cow[i].t);            cow[i].local=i;        }        sort(cow,cow+n,cmp);        Q.push(cow[0]);        int cnt=1;        ans[ cow[0].local ]=cnt;        FRL(i,1,n)        {            if (!Q.empty()&&Q.top().t<cow[i].s)            {                st=Q.top();                Q.pop();                ans[ cow[i].local ]=ans[ st.local ];            }            else                ans[ cow[i].local ]=++cnt;            Q.push(cow[i]);        }        printf("%d\n",cnt);        FRL(i,0,n)        pf("%d\n",ans[i]);    }    return 0;}/*51 102 43 65 84 7*/