POJ3622 Gourmet Grazers（FHQ Treap）
POJ3622 Gourmet Grazers（FHQ Treap）
Description
Like so many others, the cows have developed very haughty tastes and will no longer graze on just any grass. Instead, Farmer John must purchase gourmet organic grass at the Green Grass Grocers store for each of his N (1 ≤ N ≤ 100,000) cows.
Each cow i demands grass of price at least Ai (1 ≤ Ai ≤ 1,000,000,000) and with a greenness score at least Bi (1 ≤ Bi ≤ 1,000,000,000). The GGG store has M (1 ≤ M ≤ 100,000) different types of grass available, each with a price Ci (1 ≤ Ci ≤ 1,000,000,000) and a greenness score of Di (1 ≤ Di ≤ 1,000,000,000). Of course, no cow would sacrifice her individuality, so no two cows can have the same kind of grass.
Help Farmer John satisfy the cows' expensive gourmet tastes while spending as little money as is necessary.
Input
* Line 1: Two space-separated integers: N and M. * Lines 2..N+1: Line i+1 contains two space-separated integers: Ai and Bi * Lines N+2..N+M+1: Line i+N+1 contains two space-separated integers: Ci and Di
Output
* Line 1: A single integer which is the minimum cost to satisfy all the cows. If that is not possible, output -1.
Sample Input
4 7
1 1
2 3
1 4
4 2
3 2
2 1
4 3
5 2
5 4
2 6
4 4Sample Output
12Source
USACO 2007 December Gold
一开始一眼二分图，但是时间复杂度太高了
我们考虑贪心
因为他让着求最小的钱数
所以我们按照钱数排序，
然后枚举牧草，对于每一个牧草，我们统计一下它可以满足哪些奶牛
然后在能满足的奶牛中取一个需要的新鲜度离他最近的
这个过程显然可以用平衡树维护。
平衡树我用的是FHQ Treap
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<ctime>
5 #include<cstdlib>
6 #include<algorithm>
7 using namespace std;
8 #define ls T[now].ch[0]
9 #define rs T[now].ch[1]
10 const int MAXN=3*1e5+10;
11 inline char nc()
12 {
13 static char buf[MAXN],*p1=buf,*p2=buf;
14 return p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++;
15 }
16 inline int read()
17 {
18 char c=nc();int x=0,f=1;
19 while(c<'0'||c>'9'){if(c=='-')f=-1;c=nc();}
20 while(c>='0'&&c<='9'){x=x*10+c-'0',c=nc();}
21 return x*f;
22 }
23 struct node
24 {
25 int val,ch[2],pri,siz;
26 }T[MAXN];
27 int x,y,z,root=0,pos;
28 int tot=0;
29 void update(int now)
30 {
31 T[now].siz=T[ls].siz+T[rs].siz+1;
32 }
33 inline int newnode(int v)
34 {
35 T[++tot].val=v;
36 T[tot].pri=rand();
37 T[tot].siz=1;
38 return tot;
39 }
40 int merge(int x,int y)
41 {
42 if(!x||!y) return x+y;
43 if(T[x].pri<T[y].pri)
44 {
45 T[x].ch[1]=merge(T[x].ch[1],y);
46 update(x);
47 return x;
48 }
49 else
50 {
51 T[y].ch[0]=merge(x,T[y].ch[0]);
52 update(y);
53 return y;
54 }
55 }
56 void split(int now,int k,int &x,int &y)
57 {
58 if(!now) {x=y=0;return ;}
59 if(T[now].val<=k) x=now,split(rs,k,rs,y);
60 else y=now,split(ls,k,x,ls);
61 update(now);
62 }
63 struct N
64 {
65 int x,y;
66 }a[MAXN],b[MAXN];
67 int comp(const N &a,const N &b)
68 {
69 return a.x<b.x||(a.x==b.x&&a.y<b.y);
70 }
71 void insert(int val)
72 {
73 split(root,val,x,y);
74 root=merge( merge(x,newnode(val)) , y );
75 }
76 int kth(int now,int x)
77 {
78 while(now)
79 {
80 if(T[ls].siz+1==x) return now;
81 else if(T[ls].siz>=x) now=ls;
82 else x-=T[ls].siz+1,now=rs;
83 }
84 }
85 void Delet(int now)
86 {
87 split(root,now,x,z);
88 split(x,now-1,x,y);
89 y=merge(T[y].ch[0] , T[y].ch[1] );
90 root=merge( merge(x,y) ,z );
91 }
92 int main()
93 {
94 #ifdef WIN32
95 freopen("a.in","r",stdin);
96 #else
97 #endif
98 int n,m;
99 scanf("%d%d",&n,&m);
100 if (m<n){puts("-1");return 0;}
101 for(int i=1;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].y);
102 for(int i=1;i<=m;i++) scanf("%d%d",&b[i].x,&b[i].y);
103 sort(a+1,a+n+1,comp);
104 sort(b+1,b+m+1,comp);
105 long long int cur=1,ans=0,num=0;//在第一个中找到了几个
106 for(int i=1;i<=m;i++)
107 {
108 while(a[cur].x<=b[i].x&&cur<=n)
109 insert(a[cur].y),cur++;
110 split(root,b[i].y,x,y);
111 pos=kth(x,T[x].siz);
112 if(pos==0) continue;
113 num++;
114 root=merge(x,y);
115 Delet(T[pos].val);
116 ans+=b[i].x;
117 }
118 if(num==n) printf("%lld",ans);
119 else printf("-1");
120 return 0;
121 }