POJ3190 Stall Reservations 【贪婪】

Stall Reservations

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

5
1 10
2 4
3 6
5 8
4 7

Sample Output

4
1
2
3
2
4

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

/*
** 用堆维护的贪心题。先依照開始时间排序。再将牛依次放入堆里。放入之前更新堆顶元素。
** TLE到吐。。。
*/#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>#define maxn 50010
using namespace std;struct Node2 {
    int num, u, v;
    friend bool operator<(const Node2& a, const Node2& b) {
        return a.v > b.v;
    }
} cow[maxn];
int ans[maxn];bool cmp(const Node2& a, const Node2& b) {
    return a.u < b.u;
}int main() {
    int N, i, j, sum, u, v, flag;
    Node2 tmp;
    while(scanf("%d", &N) == 1) {
        sum = 0;
        for(i = 0; i < N; ++i) {
            scanf("%d%d", &u, &v);
            cow[i].num = i + 1;
            cow[i].u = u;
            cow[i].v = v;
            ans[i + 1] = 0;
        }
        sort(cow, cow + N, cmp);        priority_queue<Node2> PQ;
        PQ.push(cow[0]);
        ans[cow[0].num] = ++sum;        for(i = 1; i < N; ++i) {
            tmp = PQ.top();
            if(cow[i].u > tmp.v) {
                tmp.v = cow[i].v;
                ans[cow[i].num] = ans[tmp.num];
                PQ.pop(); PQ.push(tmp);
            } else {
                ans[cow[i].num] = ++sum;
                PQ.push(cow[i]);
            }
        }        printf("%d\n", sum);
        for(i = 1; i <= N; ++i)
            printf("%d\n", ans[i]);
    }
    return 0;
}

超时代码1：

#include <stdio.h>
#include <string.h>
#include <algorithm>#define maxn 50010
using std::sort;struct Node {
    int u, v;
} E[maxn];
struct Node2 {
    int num, u, v;
} cow[maxn];
int ans[maxn];bool cmp(const Node2& a, const Node2& b) {
    return a.u < b.u;
}int main() {
    int N, i, j, sum, u, v;
    while(scanf("%d", &N) == 1) {
        sum = 0;
        for(i = 0; i < N; ++i) {
            scanf("%d%d", &u, &v);
            cow[i].num = i + 1;
            cow[i].u = u;
            cow[i].v = v;
        }
        sort(cow, cow + N, cmp);        for(i = 0; i < N; ++i) {
            E[i].v = 0;
            for(j = 0; j <= i; ++j) {
                if(cow[i].u > E[j].v) {
                    if(!E[j].v) ++sum;
                    E[j].v = cow[i].v;
                    E[j].u = cow[i].u;
                    ans[cow[i].num] = j + 1;
                    break;
                }
            }
        }        printf("%d\n", sum);
        for(i = 1; i <= N; ++i)
            printf("%d\n", ans[i]);
    }
    return 0;
}

超时代码2：

#include <stdio.h>
#include <string.h>
#include <algorithm>#define maxn 50010
using std::sort;struct Node {
    int u, v;
} E[maxn];
struct Node2 {
    int num, u, v;
} cow[maxn];
int ans[maxn];bool cmp(const Node2& a, const Node2& b) {
    return a.u < b.u;
}int main() {
    int N, i, j, sum, u, v, flag;
    while(scanf("%d", &N) == 1) {
        sum = 0;
        for(i = 0; i < N; ++i) {
            scanf("%d%d", &u, &v);
            cow[i].num = i + 1;
            cow[i].u = u;
            cow[i].v = v;
            ans[i + 1] = 0;
        }
        sort(cow, cow + N, cmp);        for(i = 0; i < N; ++i) {
            if(ans[cow[i].num]) continue;
            ans[cow[i].num] = ++sum;
            flag = cow[i].v;
            for(j = i + 1; j < N; ++j)
                if(!ans[cow[j].num] && cow[j].u > flag) {
                    flag = cow[j].v;
                    ans[cow[j].num] = sum;
                }
        }        printf("%d\n", sum);
        for(i = 1; i <= N; ++i)
            printf("%d\n", ans[i]);
    }
    return 0;
}

版权声明：本文博客原创文章，博客，未经同意，不得转载。