usaco Milking Cows
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一题枚举,比较好实现的方法是,sort开始的时间。在枚举的时候,用pre和last维护区间。
那么最长的挤奶时间是ans1 = max(ans1,last - pre);
间隔时间在枚举的时候很好找,只要下一头奶牛的开始时间大于last便有间隔(已sort)
/** PROG:milk2 LANG:C++ ID:DickensTone**/#include<iostream>#include<fstream>#include<cstring>#include<algorithm>using namespace std;const int maxn = 5000 + 5;struct node{ int beg; int en;}person[maxn];bool cmp(node x, node y){ if(x.beg < y.beg) return 1; else if(x.beg == y.beg) return x.en < y.en; else return 0;}int main(){ freopen("milk2.in", "r", stdin); freopen("milk2.out", "w", stdout); int n; while(scanf("%d", &n) == 1) { for(int i = 0; i < n; i++) scanf("%d%d", &person[i].beg, &person[i].en); sort(person, person + n, cmp); int ans1 = 0, ans2 = 0; int pre = person[0].beg, last = person[0].en; for(int i = 1; i < n; i++) { ans1 = max(ans1, last - pre); if(person[i].beg > last) { ans2 = max(ans2, person[i].beg - last); pre = person[i].beg; last = person[i].en; } if(person[i].en > last) last = person[i].en; } ans1 = max(ans1, last - pre); printf("%d %d\n", ans1, ans2); } return 0;}
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