POJ3190：stall reservations，考点：贪心策略

原题：http://poj.org/problem?id=3190

描述

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.

中文简洁易懂型描述：

输入

Line 1: A single integer, N 

Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.

输出

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.

样例输入

5
1 10
2 4
3 6
5 8
4 7

样例输出

4
1
2
3
2
4

解法

思路：贪心

 

 注意事项：要记录奶牛原来在数组中的编号，因为输出的时候要用。以及优先队列的排序是反过来的，要记住。

 

 代码如下：（不一定AC，因为我找不到POJ账号了）

#include <iostream>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
const int MAXN = 50005;
struct cow {
    int start;
    int end;
    int index;//原来在数组中的坐标
    cow(int s,int e,int ori):start(s),end(e),index(ori){}
    bool operator <(const cow A) const {
        return start < A.start;
    }
};
struct stall {
    int id;
    int lastendtime = -1;
    stall(int l,int i):id(i),lastendtime(l){}
};
class compare {
public:
    bool operator()(const stall A, const stall B) {
        return A.lastendtime > B.lastendtime;
    }
};
int main() {
    int N;
    cin >> N;
    vector<cow>cows;
    priority_queue<stall,vector<stall>,compare>stalls;
    int result[MAXN];
    for (int i = 0; i < N; i++) {
        int s, e;
        cin >> s >> e;
        cows.push_back(cow(s, e, i));
    }
    sort(cows.begin(), cows.end());
    int stallnum = 1;
    stall top=stall(cows[0].end,stallnum++);
    result[cows[0].index] = top.id;
    stalls.push(top);
    for (int i = 1; i < N;i++) {
        top = stalls.top();//最早结束的
        if (top.lastendtime < cows[i].start) {
            stalls.pop();
            top.lastendtime = cows[i].end;
            result[cows[i].index] = top.id;
            stalls.push(top);
        }
        else {
            stall temp = stall(cows[i].end, stallnum++);
            result[cows[i].index] = temp.id;
            stalls.push(temp);
        }
    }
    cout << stalls.size() << endl;
    for (int i = 0; i < N; i++)
        cout << result[i] << endl;
    return 0;
}