1672: [Usaco2005 Dec]Cleaning Shifts 清理牛棚
Description
Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most obliging of farmers, has no choice but hire some of the cows to clean the barn. Farmer John has N (1 <= N <= 10,000) cows who are willing to do some cleaning. Because dust falls continuously, the cows require that the farm be continuously cleaned during the workday, which runs from second number M to second number E during the day (0 <= M <= E <= 86,399). Note that the total number of seconds during which cleaning is to take place is E-M+1. During any given second M..E, at least one cow must be cleaning. Each cow has submitted a job application indicating her willingness to work during a certain interval T1..T2 (where M <= T1 <= T2 <= E) for a certain salary of S (where 0 <= S <= 500,000). Note that a cow who indicated the interval 10..20 would work for 11 seconds, not 10. Farmer John must either accept or reject each individual application; he may NOT ask a cow to work only a fraction of the time it indicated and receive a corresponding fraction of the salary. Find a schedule in which every second of the workday is covered by at least one cow and which minimizes the total salary that goes to the cows.
Input
* Line 1: Three space-separated integers: N, M, and E. * Lines 2..N+1: Line i+1 describes cow i's schedule with three space-separated integers: T1, T2, and S.
Output
* Line 1: a single integer that is either the minimum total salary to get the barn cleaned or else -1 if it is impossible to clean the barn.
Sample Input
0 2 3 //一号牛,从0号stall打扫到2号,工资为3
3 4 2
0 0 1
INPUT DETAILS:
FJ has three cows, and the barn needs to be cleaned from second 0 to second
4. The first cow is willing to work during seconds 0, 1, and 2 for a total
salary of 3, etc.
Sample Output
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #define MAX 100010 7 using namespace std; 8 int n,m,e; 9 struct cow{ 10 int begin,end,cost; 11 }a[MAX]; 12 struct tree{ 13 int l,r,sum,lazy,mi; 14 }f[MAX*4]; 15 16 bool cmp(const cow&a,const cow&b){ 17 return a.begin<b.begin; 18 } 19 20 void pushup(int i){ 21 f[i].mi=min(f[i*2+1].mi,f[i*2].mi); 22 } 23 24 void update(int i,int x) 25 { 26 f[i].mi=min(f[i].mi,x); 27 f[i].lazy=min(f[i].lazy,x); 28 return; 29 } 30 31 void pushdown(int i){ 32 if(f[i].lazy!=99999999){ 33 update(i*2,f[i].lazy); 34 update(i*2+1,f[i].lazy); 35 f[i].lazy=99999999; 36 return; 37 } 38 else return; 39 } 40 41 void build(int i,int left,int right){ 42 int mid=(left+right)/2; 43 f[i].lazy=99999999;f[i].l=left;f[i].r=right;f[i].mi=99999999; 44 if(left==right){ 45 f[i].mi=99999999; 46 f[i].lazy=99999999; 47 return; 48 } 49 build(i*2,left,mid); 50 build(i*2+1,mid+1,right); 51 pushup(i); 52 } 53 54 void add(int i,int left,int right,int v){ 55 int mid=(f[i].l+f[i].r)/2; 56 if(f[i].l==left&&f[i].r==right){ 57 update(i,v); 58 return; 59 } 60 pushdown(i); 61 if(mid>=right)add(i*2,left,right,v); 62 else if(mid<left)add(i*2+1,left,right,v); 63 else add(i*2,left,mid,v),add(i*2+1,mid+1,right,v); 64 pushup(i); 65 } 66 67 int MIN(int i,int left,int right){ 68 int mid=(f[i].l+f[i].r)/2; 69 if(left<m) return 0; 70 if(f[i].l==left&&f[i].r==right) return f[i].mi; 71 pushdown(i); 72 if(mid>=right) return MIN(i*2,left,right); 73 if(mid<left) return MIN(i*2+1,left,right); 74 return min(MIN(i*2,left,mid),MIN(i*2+1,mid+1,right)); 75 } 76 int main(){ 77 scanf("%d%d%d",&n,&m,&e); 78 build(1,m,e); 79 for(int i=1;i<=n;i++) scanf("%d%d%d",&a[i].begin,&a[i].end,&a[i].cost); 80 sort(a+1,a+n+1,cmp); 81 for(int i=1;i<=n;i++){ 82 if(MIN(1,a[i].begin-1,a[i].begin-1)==99999999){ 83 printf("%d",-1); 84 return 0; 85 } 86 add(1,a[i].begin,a[i].end,(MIN(1,a[i].begin-1,a[i].begin-1)+a[i].cost)); 87 } 88 int ans=MIN(1,e,e); 89 if(ans==99999999)printf("%d",-1); 90 else printf("%d",ans); 91 return 0; 92 }