BZOJ1672: [Usaco2005 Dec]Cleaning Shifts 清理牛棚

标签：des style blog http color io os ar for

1672: [Usaco2005 Dec]Cleaning Shifts 清理牛棚

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 414 Solved: 191
[Submit][Status]

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.

约翰的奶牛们从小娇生惯养，她们无法容忍牛棚里的任何脏东西．约翰发现，如果要使这群有洁癖的奶牛满意，他不得不雇佣她们中的一些来清扫牛棚, 约翰的奶牛中有N(1≤N≤10000)头愿意通过清扫牛棚来挣一些零花钱．由于在某个时段中奶牛们会在牛棚里随时随地地乱扔垃圾，自然地，她们要求在这段时间里，无论什么时候至少要有一头奶牛正在打扫．需要打扫的时段从某一天的第M秒开始，到第E秒结束f0≤M≤E≤86399)．注意这里的秒是指时间段而不是时间点，也就是说，每天需要打扫的总时间是E-M+I秒．约翰已经从每头牛那里得到了她们愿意接受的工作计划：对于某一头牛，她每天都愿意在笫 Ti，.T2秒的时间段内工作(M≤Ti≤马≤E)，所要求的报酬是S美元(0≤S≤500000)．与需打扫时段的描述一样，如果一头奶牛愿意工作的时段是每天的第10_20秒，那她总共工作的时间是11秒，而不是10秒．约翰一旦决定雇佣某一头奶牛，就必须付给她全额的工资，而不能只让她工作一段时间，然后再按这段时间在她愿意工作的总时间中所占的百分比来决定她的工资．现在请你帮约翰决定该雇佣哪些奶牛以保持牛棚的清洁，当然，在能让奶牛们满意的前提下，约翰希望使总花费尽量小．

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.

第1行：3个正整数N，M，E，用空格隔开．

第2到N+1行：第i+l行给出了编号为i的奶牛的工作计划，即3个用空格隔开的正整数Ti，T2，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.

输出一个整数，表示约翰需要为牛棚清理工作支付的最少费用．如果清理工作不可能完成，

那么输出-1.

Sample Input

3 0 4 //三头牛,要打扫从0到4号stall
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

HINT

约翰有3头牛，牛棚在第0秒到第4秒之间需要打扫．第1头牛想要在第0，1，2秒内工作，为此她要求的报酬是3美元．其余的依此类推．约翰雇佣前两头牛清扫牛棚，可以只花5美元就完成一整天的清扫．

Source

Silver

题解：

这道题不错，线段树优化DP。

按照l端点排序牛

f[i]表示到i时刻的最小代价，那么对于一只l-r费用c的牛

可以用f[i-1]+c更新l-r

用线段树维护，单点查询，区间更新

代码：

（不知道哪里写残了T_T）

  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 100000000000
 24 
 25 #define maxn 80000+2000
 26 
 27 #define maxm 500+100
 28 
 29 #define eps 1e-10
 30 
 31 #define ll long long
 32 
 33 #define pa pair<int,int>
 34 
 35 #define for0(i,n) for(int i=0;i<=(n);i++)
 36 
 37 #define for1(i,n) for(int i=1;i<=(n);i++)
 38 
 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 40 
 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 42 
 43 #define mod 1000000007
 44 
 45 using namespace std;
 46 
 47 inline int read()
 48 
 49 {
 50 
 51     int x=0,f=1;char ch=getchar();
 52 
 53     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 54 
 55     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
 56 
 57     return x*f;
 58 
 59 }
 60 
 61 struct rec{int l,r;ll w;}a[maxn];
 62 
 63 struct seg{int l,r;ll mi,tag;}t[4*maxn];
 64 
 65 int n,x,y;
 66 
 67 inline bool cmp(rec a,rec b)
 68 
 69 {
 70 
 71     return a.l<b.l||(a.l==b.l&&a.r<b.r);
 72 
 73 }
 74 
 75 inline void pushup(int k)
 76 
 77 {
 78 
 79     t[k].mi=min(t[k<<1].mi,t[k<<1|1].mi);
 80 
 81 }
 82 
 83 inline void update(int k,ll z)
 84 
 85 {
 86 
 87     t[k].mi=min(t[k].mi,z);
 88 
 89     t[k].tag=min(t[k].tag,z);
 90 
 91 }
 92 
 93 inline void pushdown(int k)
 94 
 95 {
 96 
 97     if(t[k].tag==-1)return ;
 98 
 99     update(k<<1,t[k].tag);
100 
101     update(k<<1|1,t[k].tag);
102 
103     t[k].tag=-1;
104 
105 }
106 
107 inline void build(int k,int x,int y)
108 
109 {
110 
111     int l=t[k].l=x,r=t[k].r=y,mid=(l+r)>>1;t[k].tag=-1;
112 
113     if(l==r){t[k].mi=l==0?0:inf;return ;}
114 
115     build(k<<1,l,mid);build(k<<1|1,mid+1,r);
116 
117     pushup(k);
118 
119 }
120 
121 inline ll query(int k,int x)
122 
123 {
124 
125     int l=t[k].l,r=t[k].r,mid=(l+r)>>1;
126 
127     if(l==r)return t[k].mi;
128 
129     pushdown(k);
130 
131     if(x<=mid)return query(k<<1,x);else return query(k<<1|1,x);
132 
133 }
134 
135 inline void change(int k,int x,int y,ll z)
136 
137 {
138 
139     int l=t[k].l,r=t[k].r,mid=(l+r)>>1;
140 
141     if(l==x&&r==y){update(k,z);return;}
142 
143     pushdown(k);
144 
145     if(y<=mid)change(k<<1,x,y,z);
146 
147     else if(x>mid)change(k<<1|1,x,y,z);
148 
149     else change(k<<1,x,mid,z),change(k<<1|1,mid+1,y,z);
150 
151     pushup(k);
152 
153 }
154 
155 int main()
156 
157 {
158 
159     freopen("input.txt","r",stdin);
160 
161     freopen("output.txt","w",stdout);
162 
163     n=read();x=read();y=read();
164 
165     for1(i,n)
166 
167      {
168 
169       a[i].l=read()-x+1;a[i].r=read()-x+1;a[i].w=read();
170 
171       if(a[i].l<=0)a[i].l=1;
172 
173      }
174 
175     sort(a+1,a+n+1,cmp);
176 
177     build(1,0,y-x+1);
178 
179     for1(i,n)
180 
181     {
182 
183         if(a[i].r<a[i].l)continue;
184 
185         if(a[i].r<=0)continue;
186 
187         if(a[i].l>y-x+1)continue;
188 
189         ll t=query(1,a[i].l-1);
190 
191         if(t!=inf)change(1,a[i].l,min(a[i].r,y-x+1),t+a[i].w);else break;
192 
193     }
194 
195     //for1(i,n)cout<<a[i].l<<‘ ‘<<a[i].r<<‘ ‘<<a[i].w<<endl;
196 
197     //for1(k,8*n)cout<<t[k].l<<‘ ‘<<t[k].r<<‘ ‘<<t[k].mi<<‘ ‘<<t[k].tag<<endl;
198 
199     ll ans=query(1,y-x+1);
200 
201     printf("%lld\n",ans==inf?-1:ans);
202 
203     return 0;
204 
205 }

View Code

hzwer的

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #define ll long long
 7 #define inf 10000000000
 8 using namespace std;
 9 inline ll read()
10 {
11     int x=0,f=1;char ch=getchar();
12     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
13     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
14     return x*f;
15 }
16 int n,bg,ed;
17 struct seg{int l,r;ll tag,mn;}t[400005];
18 struct data{int t1,t2,c;}a[10005];
19 inline bool operator<(data a,data b)
20 {
21     return a.t1<b.t1;
22 }
23 void pushdown(int k)
24 {
25     if(t[k].l==t[k].r)return;
26     ll tag=t[k].tag;t[k].tag=inf;
27     if(tag!=inf)
28     {
29         t[k<<1].tag=min(t[k<<1].tag,tag);
30         t[k<<1|1].tag=min(t[k<<1|1].tag,tag);
31         t[k<<1].mn=min(t[k<<1].mn,tag);
32         t[k<<1|1].mn=min(t[k<<1|1].mn,tag);
33     }
34 }
35 void build(int k,int l,int r)
36 {
37     t[k].l=l;t[k].r=r;t[k].mn=inf;t[k].tag=inf;
38     if(l==r)return;
39     int mid=(l+r)>>1;
40     build(k<<1,l,mid);build(k<<1|1,mid+1,r);
41 }
42 ll query(int k,int x)
43 {
44     pushdown(k);
45     if(x<bg)return 0;
46     int l=t[k].l,r=t[k].r;
47     if(l==r)return t[k].mn;
48     int mid=(l+r)>>1;
49     if(x<=mid)return query(k<<1,x);
50     else return query(k<<1|1,x);
51 }
52 void update(int k,int x,int y,ll val)
53 {
54     pushdown(k);
55     int l=t[k].l,r=t[k].r;
56     if(x==l&&y==r)
57     {
58         t[k].tag=val;
59         t[k].mn=min(t[k].mn,val);
60         return;
61     }
62     int mid=(l+r)>>1;
63     if(y<=mid)update(k<<1,x,y,val);
64     else if(x>mid)update(k<<1|1,x,y,val);
65     else
66     {
67         update(k<<1,x,mid,val);
68         update(k<<1|1,mid+1,y,val);
69     }
70 }
71 int main()
72 {
73     n=read();bg=read();ed=read();
74     build(1,bg,ed);
75     for(int i=1;i<=n;i++)
76         a[i].t1=read(),a[i].t2=read(),a[i].c=read();
77     sort(a+1,a+n+1);
78     for(int i=1;i<=n;i++)
79     {
80         int t1=a[i].t1,t2=a[i].t2,c=a[i].c;
81         ll tmp=query(1,t1-1);
82         if(tmp==inf){puts("-1");return 0;}
83         update(1,t1,t2,tmp+c);
84     }
85     ll ans=query(1,ed);
86     if(ans<inf)printf("%lld",ans);
87     else puts("-1");
88     return 0;
89 }

View Code

BZOJ1672: [Usaco2005 Dec]Cleaning Shifts 清理牛棚

标签：des style blog http color io os ar for

原文地址：http://www.cnblogs.com/zyfzyf/p/3992586.html