POJ3190 Stall Reservations 贪心

来源：互联网发布：cf英雄武器抽奖软件编辑：程序博客网时间：2024/05/16 11:58

这题用贪心法。

首先对每头牛挤奶的开始时间进行升序排序，然后假设已经使用的stall的集合为S，初始时S为空，每个stall要记录下一个T，表示在T时刻后，该stall变为空闲。

接着，按顺序对每头奶牛进行查询，在S中选择最小的T的stall，如果T小于该头奶牛的开始挤奶时间，则将此stall分配给该奶牛，同时更新T为该奶牛的结束时间；如果找不到满足条件的stall，就意味着要再申请一个stall了。

#define HEAD#include <stdio.h>#include <vector>#include <math.h>#include <string.h>#include <string>#include <iostream>#include <queue>#include <list>#include <algorithm>#include <stack>#include <map>#include<iostream>  #include<cstdio>  using namespace std;struct STALL{int number;int endtime;bool operator()(const STALL& a, const STALL& b){return a.endtime > b.endtime;}};struct COW{int start;int end;int pos;int number;};bool COWComp(COW& a1, COW& a2){return a1.start < a2.start;}bool COWcomp2(COW& a1, COW& a2){return a1.number < a2.number;}int COSPOS[50001];int main(){#ifdef _DEBUGfreopen("d:\\in.txt", "r", stdin);#endifmemset(COSPOS, 0, sizeof(COSPOS));int n;scanf("%d\n", &n);priority_queue<STALL, vector<STALL>, STALL> pq;vector<COW> cows;for (int i = 0; i < n;i++){COW cow;scanf("%d %d\n", &cow.start, &cow.end);cow.number = i;cows.push_back(cow);}sort(cows.begin(), cows.end(), COWComp);int stallnumber = 0;for (int i = 0; i < n;i++){if (pq.size() == 0 || pq.top().endtime >= cows[i].start){STALL newstall;newstall.number = ++stallnumber;newstall.endtime = cows[i].end;cows[i].pos = stallnumber;pq.push(newstall);}else if (pq.top().endtime < cows[i].start){STALL stall = pq.top();pq.pop();stall.endtime = cows[i].end;pq.push(stall);cows[i].pos = stall.number;}}printf("%d\n", stallnumber);sort(cows.begin(), cows.end(), COWcomp2);for (int i = 0; i < n; i++){printf("%d\n", cows[i].pos);}return 0;}

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