Cleaning Shifts
Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000), the first being shift 1 and the last being shift T.
Each cow is only available at some interval of times during the day for work on cleaning. Any cow that is selected for cleaning duty will work for the entirety of her interval.
Your job is to help Farmer John assign some cows to shifts so that (i) every shift has at least one cow assigned to it, and (ii) as few cows as possible are involved in cleaning. If it is not possible to assign a cow to each shift, print -1.
Input
* Line 1: Two space-separated integers: N and T
* Lines 2..N+1: Each line contains the start and end times of the interval during which a cow can work. A cow starts work at the start time and finishes after the end time.
Output
* Line 1: The minimum number of cows Farmer John needs to hire or -1 if it is not possible to assign a cow to each shift.
Sample Input
3 10
1 7
3 6
6 10
Sample Output
2
Hint
This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.
INPUT DETAILS:
There are 3 cows and 10 shifts. Cow #1 can work shifts 1..7, cow #2 can work shifts 3..6, and cow #3 can work shifts 6..10.
OUTPUT DETAILS:
By selecting cows #1 and #3, all shifts are covered. There is no way to cover all the shifts using fewer than 2 cows.
Source
题目类型：排序+贪心
算法分析：首先按照开始时间递增排序，然后按照结束时间递增排序，由贪心策略可知一定要选择第一个区间，且如果这个区间的左端点大于1，则直接输出-1，否则每次向右找满足x < tt.y + 1的最大右端点的值。然后看最后是否能够将t这个位置覆盖住
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#pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define LL long long #define ULL unsigned long long #define DB(ccc) cout << #ccc << " = " << ccc << endl const int INF = 0x7FFFFFFF; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e6 + 66; struct node{int x, y;}; node aa[maxn]; bool operator < (const node &a, const node &b) { if (a.x != b.x) return a.x < b.x; else return a.y < b.y; } int main() { // CFF; int n, t; while (scanf ("%d%d", &n, &t) != EOF) { for (int i = 1; i <= n; i++) scanf ("%d%d", &aa[i].x, &aa[i].y); sort (aa + 1, aa + 1 + n); int pos = 0, tmax, res = 0, i = 1; while (i <= n && pos < t) { res++; tmax = pos; if (aa[i].x > pos + 1) { puts ("-1"); return 0; } while (i <= n && aa[i].x <= pos + 1) { tmax = max (tmax, aa[i].y); i++; } pos = tmax; } if (pos >= t) printf ("%d\n", res); else puts ("-1"); } return 0; } |
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