• # poj2376 Cleaning Shifts【线段树】【DP】

Cleaning Shifts
 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 32561 Accepted: 7972

Description

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

https://www.cnblogs.com/wyboooo/p/9808378.html

和poj3171基本一样，改一下输入和范围即可。

```  1 #include <iostream>
2 #include <set>
3 #include <cmath>
4 #include <stdio.h>
5 #include <cstring>
6 #include <algorithm>
7 using namespace std;
8 typedef long long LL;
9 #define inf 0x7f7f7f7f
10
11 const int maxn = 25000 + 5;
12 const int maxtime = 1e6 + 5;
13 struct node{
14     int st, ed, cost;
15 }cow[maxn];
16 bool cmp(node a, node b)
17 {
18     return a.ed < b.ed;
19 }
20 LL tree[maxtime << 2];//区间中f[]最小值
21 int n, L, R;
22
23 void pushup(int rt)
24 {
25     tree[rt] = min(tree[rt << 1], tree[rt << 1|1]);
26 }
27
28 void build(int rt, int l, int r)
29 {
30     if(l == r){
31         tree[maxn] = inf;
32         return;
33     }
34     int mid = (l + r) / 2;
35     build(rt<<1, l, mid);
36     build(rt<<1|1, mid + 1, r);
37     pushup(rt);
38 }
39
40 void update(int x, LL val, int l, int r, int rt)
41 {
42     if(l == r){
43         tree[rt] = min(tree[rt], val);
44         return;
45     }
46     int m = (l + r) / 2;
47     if(x <= m){
48         update(x, val, l, m, rt<<1);
49     }
50     else{
51         update(x, val, m + 1, r, rt<<1|1);
52     }
53     pushup(rt);
54 }
55
56 LL query(int L, int R, int l, int r, int rt)
57 {
58     if(L <= l && R >= r){
59         return tree[rt];
60     }
61     int m = (l + r) / 2;
62     LL ans = inf;
63     if(L <= m){
64         ans = min(ans, query(L, R, l, m, rt<< 1));
65     }
66     if(R > m){
67         ans = min(ans, query(L, R, m + 1, r, rt<<1|1));
68     }
69     pushup(rt);
70     return ans;
71 }
72
73 int main()
74 {
75     while(scanf("%d%d", &n, &R) != EOF){
76         R+=1;
77         memset(tree, 0x7f, sizeof(tree));
78         for(int i = 1; i <= n; i++){
79             scanf("%d%d", &cow[i].st, &cow[i].ed);
80             cow[i].st+=1;cow[i].ed+=1;
81             cow[i].cost = 1;
82         }
83         sort(cow + 1, cow + 1 + n, cmp);
84
85         build(1, 1, R);
86
87         update(1, 0, 1, R, 1);
88         //cout<<"yes"<<endl;
89         //int far = L;
90         bool flag = true;
91         for(int i = 1; i <= n; i++){
92             /*if(cow[i].st > far + 1){
93                 flag = false;
94             //    break;
95             }*/
96             int a = max(1, cow[i].st - 1);
97             int b = min(R, cow[i].ed);
98             //cout<<a<<" "<<b<<endl;
99             LL f = query(a, b, 1, R, 1);
100             f += cow[i].cost;
101             //cout<<f<<endl;
102             update(b, f, 1, R, 1);
103             //far = max(far, cow[i].ed);
104             //cout<<far<<endl;
105         }
106         //cout<<"yes"<<endl;
107
108         LL ans = query(R, R, 1, R, 1);
109         if(ans >= inf){
110             printf("-1
");
111         }
112         else{
113                 printf("%lld
", ans);
114
115         //else{
116         //    printf("-1
");
117         }
118
119     }
120
121 }```
• 相关阅读:
Shuffle Cards
求和VII
Finite Encyclopedia of Integer Sequences(找规律)
Codeforces Round #223 (Div. 2) C
Codeforces Round #223 (Div. 2) A
题目1047：素数判定
Codeforces Round #219 (Div. 1) C. Watching Fireworks is Fun
Codeforces Round #219 (Div. 2) B. Making Sequences is Fun
中南大学第一届长沙地区程序设计邀请赛 New Sorting Algorithm