【USACO1.2.1】Milking Cows

Milking Cows

Three farmers rise at 5 am each morning and head for the barn to milk three cows. The first farmer begins milking his cow at time 300 (measured in seconds after 5 am) and ends at time 1000. The second farmer begins
at time 700 and ends at time 1200. The third farmer begins at time 1500 and ends at time 2100. The longest continuous time during which at least one farmer was milking a cow was 900 seconds (from 300 to 1200). The longest time no milking was done, between
the beginning and the ending of all milking, was 300 seconds (1500 minus 1200).

Your job is to write a program that will examine a list of beginning and ending times for N (1 <= N <= 5000) farmers milking N cows and compute (in seconds):

• The longest time interval at least one cow was milked.
• The longest time interval (after milking starts) during which no cows were being milked.

PROGRAM NAME: milk2

INPUT FORMAT

SAMPLE INPUT (file milk2.in)

3
300 1000
700 1200
1500 2100

OUTPUT FORMAT

A single line with two integers that represent the longest continuous time of milking and the longest idle time.

SAMPLE OUTPUT (file milk2.out)

900 300

懒得翻译直接搬救兵 http://www.nocow.cn/index.php/Translate:USACO/milk2

这道题才是错的最多的 因为各种各样的考虑不周到错了很多次 时间比较久了我记不清ＷＡ的地方了。。。直接上代码吧反正就是模拟

/*
ID: Alpha Augur
PROG: milk2
LANG: C++
*/
#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
using namespace std;

const int maxn = 5010;
struct Ti{
    int st, en;
}ti[maxn];

bool cmp(Ti a, Ti b);

int main(){
    freopen("milk2.in", "r", stdin);
    freopen("milk2.out", "w", stdout);
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; ++i){
        scanf("%d%d", &ti[i].st, &ti[i].en);
    }
    sort(ti, ti+n, cmp);

    int cnt = ti[0].en - ti[0].st;
    int ma = cnt;
    int mi = 0;
    for(int i = 1; i < n; ++i){
        if(ti[i].st <= ti[i-1].en){
            if(ti[i].en > ti[i-1].en){
                cnt += ti[i].en - ti[i-1].en;
                ma = max(ma, cnt);
            }
            ti[i].en = max(ti[i].en, ti[i-1].en);
        }
        else{
            cnt = ti[i].en - ti[i].st;
            ti[i].en = max(ti[i].en, ti[i-1].en);
            mi = max(ti[i].st - ti[i-1].en, mi);
        }
    }

    printf("%d %d\n", ma, mi);
    return 0;
}

bool cmp(Ti a, Ti b){
    if(a.st == b.st){
        return a.en < b.en;
    }
    return a.st < b.st;
}

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