POJ3190：Stall Reservations（贪心）

Time Limit: 1000MS
Memory Limit: 65536K
Special Judge

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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:

Other outputs using the same number of stalls are possible.

Source

USACO 2006 February Silver

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解题分析

所有奶牛都必须挤奶。到了一个奶牛的挤奶开始时间，就必须为这个奶牛找畜栏。因此按照奶牛的开始时间逐个处理它们，是必然的。

S(x)表示奶牛x的开始时间，E(x)表示x的结束时间。对E(x)，x可以是奶牛，也可以是畜栏。畜栏的结束时间，就是正在其里面挤奶的奶牛的结束时间。同一个畜栏的结束时间是不断在变的。

1)把所有奶牛按开始时间从小到大排序。
2)为第一头奶牛分配一个畜栏。
3)依次处理后面每头奶牛i。处理i时，考虑已分配畜栏中，结束时间最早的畜栏x。

• 若E(x)<S(i)，则不用分配新畜栏，i可进入x，并修改E(x)为E(i)。
• 若E(x)>=S(i)，则分配新畜栏y，记E(y)=E(i)。

直到所有奶牛处理结束。

需要用优先队列存放已经分配的畜栏，并使得结束时间最早的畜栏始终位于队列头部。

证明：
由于按开始时间的顺序处理奶牛是必然，且按该算法，为奶牛i分配新畜栏时，确实是不得不分配的，所以算法正确。

复杂度：
O(nlogn)

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AC代码

#include<iostream>#include<cstdio>#include<algorithm>#include<queue>using namespace std;struct Cow{    int a,b;    int No;    bool operator < (const Cow &c)    const{return a < c.a;}}cows[50100];int pos[50100];struct Stall{    int et;    int No;    bool operator < (const Stall &s)    const{return et > s.et;}    Stall(int e,int n):et(e),No(n){}};priority_queue<Stall>pq;int main(){    int N;    scanf("%d",&N);    for(int i=0;i<N;i++){        scanf("%d%d",&cows[i].a,&cows[i].b);        cows[i].No=i;    }    sort(cows,cows+N);    int total=0;    for(int i=0;i<N;i++){        if(pq.empty()){            total++;            pq.push(Stall(cows[i].b,total));            pos[cows[i].No]=total;        }        else{            Stall st=pq.top();            if(st.et<cows[i].a){                pq.pop();                pq.push(Stall(cows[i].b,st.No));                pos[cows[i].No]=st.No;            }            else{                total++;                pq.push(Stall(cows[i].b,total));                pos[cows[i].No]=total;            }        }    }    printf("%d\n",total);    for(int i=0;i<N;i++)        printf("%d\n",pos[i]);    return 0;}