poj 3190 Stall Reservations

Stall Reservations

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6163 Accepted: 2250 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

2 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.

每一只奶牛要求在时间区间[A, B]内独享一个牛栏。问最少需要多少个牛栏。

贪心策略是优先满足A最小的奶牛，维持一个牛栏B最小堆，将新来的奶牛塞进B最小的牛栏里。

问题的麻烦点在于如何求出每个区间是第几个牛栏。其实只需要保存每个区间是第几个，再保存放在是第几个牛栏。最后再用一次排序就可以了。

#include <iostream>#include <queue>#include <cstdio>#include <algorithm>#include <vector>using namespace std;const int maxn = 1e6 + 4;struct node{    int a, b, c, d;};node no[maxn];bool cmp(node x, node y) {    if(x.a == y.a) {        return x.b < y.b;    } else {        return x.a < y.a;    }}bool cmp2(node x, node y) {    return x.c < y.c;}struct node2 {    int x, y;    bool operator < (const node2 &a) const {        return x>a.x;//最小值优先    }};int main(){    int n;    priority_queue<node2> que;//最小值优先;    scanf("%d", &n);    for(int i = 0; i < n; ++i) {        scanf("%d%d", &no[i].a, &no[i].b);        no[i].c = i;    }    sort(no, no + n, cmp);    que.push(node2{no[0].b, 1});    int ans = 1;    no[0].d = 1;    for(int i = 1; i < n; ++i) {        node2 temp = que.top();        int tempx = temp.x;        int tempy = temp.y;        if(no[i].a > tempx) {            que.pop();            que.push(node2{no[i].b, tempy});            no[i].d = tempy;        } else {            ans++;            que.push(node2{no[i].b, ans});            no[i].d = ans;        }    }    sort(no, no + n, cmp2);    printf("%d\n", ans);    for(int i = 0; i < n; ++i) {        printf("%d\n", no[i].d);    }    return 0;}