雅礼中学第四场20180109

T1

　　就是左端点排序后，然后如果可以到达下面一个点，那么对于

　　

　　看这个，f[i]表示到了i这个点的方案数，当前边覆盖到了a[i].r

　　a[i].l-1  ------  a[i].r-1都可以转移过来，

　　然后对于其后面的来说，当前的边选不选无所谓，a[i].r---m都乘2

  1 #include<cstring>
  2 #include<algorithm>
  3 #include<cmath>
  4 #include<iostream>
  5 #include<cstdio>
  6 
  7 #define ls p<<1
  8 #define rs p<<1|1
  9 #define mod 1000000009
 10 #define ll long long
 11 #define N 500007
 12 using namespace std;
 13 inline int read()
 14 {
 15     int x=0,f=1;char ch=getchar();
 16     while(ch<'0'||ch>'9'){if (ch=='-')f=-1;ch=getchar();}
 17     while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
 18     return x*f;
 19 }
 20 
 21 int n,m;
 22 int b[N];
 23 struct Node
 24 {
 25     int l,r;
 26 }a[N];
 27 struct Date
 28 {
 29     ll add,cf,sum;
 30 }tr[N*4];
 31 
 32 bool operator<(Node x,Node y){return x.l<y.l;}
 33 void build(int p,int l,int r)
 34 {
 35     tr[p].cf=1;
 36     if (l==r)return;
 37     int mid=(l+r)>>1;
 38     build(p<<1,l,mid),build(p<<1|1,mid+1,r);
 39 }
 40 void downdate(int p)
 41 {
 42     ll cf=tr[p].cf,add=tr[p].add;tr[p].cf=1,tr[p].add=0;
 43     (tr[ls].cf*=cf)%=mod,(tr[ls].add*=cf)%=mod,(tr[ls].add+=add)%=mod;
 44     (tr[rs].cf*=cf)%=mod,(tr[rs].add*=cf)%=mod,(tr[rs].add+=add)%=mod;
 45     (tr[ls].sum*=cf)%=mod,(tr[ls].sum+=add)%=mod;
 46     (tr[rs].sum*=cf)%=mod,(tr[rs].sum+=add)%=mod;
 47 }
 48 void update(int p)
 49 {
 50     tr[p].sum=tr[ls].sum+tr[rs].sum;
 51 }
 52 void modify(int p,int l,int r,int x,int y,ll jia,ll cf)
 53 {
 54     if (l==x&&y==r)
 55     {
 56         if (jia){(tr[p].add+=jia)%=mod;(tr[p].sum+=(r-l+1)*jia)%=mod;}
 57         else
 58         {
 59             (tr[p].sum*=cf)%=mod;
 60             (tr[p].add*=cf)%=mod;
 61             (tr[p].cf*=cf)%=mod;
 62         }
 63         return;
 64     }
 65     downdate(p);
 66     int mid=(l+r)>>1;
 67     if (y<=mid) modify(p<<1,l,mid,x,y,jia,cf);
 68     else if(x>mid) modify(p<<1|1,mid+1,r,x,y,jia,cf);
 69     else modify(p<<1,l,mid,x,mid,jia,cf),modify(p<<1|1,mid+1,r,mid+1,y,jia,cf);
 70     update(p);
 71 }
 72 int query(int p,int l,int r,int x,int y)
 73 {
 74     if (l==x&&y==r) return tr[p].sum;
 75     downdate(p);
 76     int mid=(l+r)>>1;
 77     if (y<=mid) return query(p<<1,l,mid,x,y);
 78     else if (x>mid) return query(p<<1|1,mid+1,r,x,y);
 79     else return (query(p<<1,l,mid,x,mid)+query(p<<1|1,mid+1,r,mid+1,y))%mod;
 80     update(p);
 81 }
 82 int main()
 83 {
 84     n=read(),m=read();
 85     for (int i=1;i<=n;i++)a[i].l=read(),a[i].r=read();
 86     sort(a+1,a+n+1);
 87     for (int i=1;i<=m;i++)b[i]=read();
 88     sort(b+1,b+m+1);
 89     for (int i=1;i<=n;i++)
 90     {
 91         (a[i].l=lower_bound(b+1,b+m+1,a[i].l)-b);
 92         (a[i].r=upper_bound(b+1,b+m+1,a[i].r)-b)--;
 93 //        cout<<"i="<<i<<" "<<a[i].l<<" "<<a[i].r<<endl;
 94     }
 95     build(1,0,m);
 96     modify(1,0,m,0,0,1,1);
 97     int num=0;
 98     for (int i=1;i<=n;i++)
 99     {
100         if (a[i].l>a[i].r)
101         {
102             num++;
103             continue;
104         }
105         ll x=query(1,0,m,a[i].l-1,a[i].r-1);
106         modify(1,0,m,a[i].r,m,0,2);
107         modify(1,0,m,a[i].r,a[i].r,x,1);
108     }
109     ll ans=query(1,0,m,m,m);
110     for (int i=1;i<=num;i++)(ans*=(ll)2)%=mod;
111     printf("%lld\n",ans);
112 }

T2

因为x,y不同，所以一定有一位二进制不同，

但是无法区分谁大谁小

所以我们把其映射到12位上去，6个0,6个1，这样有924>920种

然后因为不同，所以找出哪一位，就是x为1，y为0的那一位，一定存在

然后就可以k=12来判断了。

代码很好实现

 

  1 // Copyright (C) 2017 __debug.
  2 
  3 // This program is free software; you can redistribute it and/or
  4 // modify it under the terms of the GNU General Public License as
  5 // published by the Free Software Foundation; version 3
  6 
  7 // This program is distributed in the hope that it will be useful,
  8 // but WITHOUT ANY WARRANTY; without even the implied warranty of
  9 // MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 10 // GNU General Public License for more details.
 11 
 12 // You should have received a copy of the GNU General Public License
 13 // along with this program; If not, see <http://www.gnu.org/licenses/>.
 14 
 15 
 16 #include <bits/stdc++.h>
 17 #include <ext/pb_ds/assoc_container.hpp>
 18 #include <ext/pb_ds/priority_queue.hpp>
 19 
 20 #define x first
 21 #define y second
 22 #define MP std::make_pair
 23 #define SZ(x) ((int)(x).size())
 24 #define ALL(x) (x).begin(), (x).end()
 25 #define DEBUG(...) fprintf(stderr, __VA_ARGS__)
 26 #ifdef __linux__
 27 #define getchar getchar_unlocked
 28 #define putchar putchar_unlocked
 29 #endif
 30 
 31 using std::pair;
 32 using std::vector;
 33 using std::string;
 34 
 35 typedef long long LL;
 36 typedef pair<int, int> Pii;
 37 
 38 const int oo = 0x3f3f3f3f;
 39 
 40 template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, true : false; }
 41 template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, true : false; }
 42 string procStatus()
 43 {
 44     std::ifstream t("/proc/self/status");
 45     return string(std::istreambuf_iterator<char>(t), std::istreambuf_iterator<char>());
 46 }
 47 template<typename T> T read(T &x)
 48 {
 49     int f = 1;
 50     char ch = getchar();
 51     for (; !isdigit(ch); ch = getchar())
 52         f = (ch == '-' ? -1 : 1);
 53     for (x = 0; isdigit(ch); ch = getchar())
 54         x = 10 * x + ch - '0';
 55     return x *= f;
 56 }
 57 template<typename T> void write(T x)
 58 {
 59     if (x == 0) {
 60         putchar('0');
 61         return;
 62     }
 63     if (x < 0) {
 64         putchar('-');
 65         x = -x;
 66     }
 67     static char s[20];
 68     int top = 0;
 69     for (; x; x /= 10)
 70         s[++top] = x % 10 + '0';
 71     while (top)
 72         putchar(s[top--]);
 73 }
 74 // EOT
 75 
 76 const int LIM = 12;
 77 
 78 int type, N;
 79 vector<int> code;
 80 
 81 void init()
 82 {
 83     for (int s = 0; s < (1 << LIM) && SZ(code) < N; ++s) {
 84         if (__builtin_popcount(s) == LIM / 2)
 85             code.push_back(s);
 86     }
 87     assert(SZ(code) == N);
 88 }
 89 
 90 int encode(int x, int y)
 91 {
 92     return 32 - __builtin_clz((code[x - 1] ^ code[y - 1]) & code[x - 1]);
 93 }
 94 
 95 bool decode(int q, int h)
 96 {
 97     return code[q - 1] >> (h - 1) & 1;
 98 }
 99 
100 int main()
101 {
102     int T;
103     read(type); read(N); read(T);
104     init();
105     while (T--) {
106         int x, y;
107         read(x); read(y);
108         if (type == 1)
109             printf("%d\n", encode(x, y));
110         else
111             puts(decode(x, y) ? "yes" : "no");
112     }
113 }
114 
115 // 鍜畾闈掑北涓嶆斁鏉撅紝绔嬫牴鍘熷湪鐮村博涓€?
116 //     -- 閮戠嚠銆婄鐭炽€?

View Code

 

T3

因为要最大化，编号不断增大，所以选了其父亲后一定会选最小的点，

但是不能直接贪心，要对于全局所有的点贪心，转化很巧妙。

代码附上，慢慢理解，

两个判断意思是，以前的原来点是没用的，并且不能是根，没有父亲。

 1 #include<cstring>
 2 #include<cmath>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<cstdio>
 6 #include<queue>
 7 
 8 #define ll long long
 9 #define N 100007
10 using namespace std;
11 inline int read()
12 {
13     int x=0,f=1;char ch=getchar();
14     while(ch<'0'||ch>'9'){if (ch=='-')f=-1;ch=getchar();}
15     while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
16     return x*f;
17 }
18 
19 struct Node
20 {
21     int u,s,t;//u鍝釜鐐癸紝s鏄潈鍊煎拰锛宼鏄痵iz銆?
22 };
23 bool operator >(Node x,Node y)
24 {
25     return (ll)x.s*y.t>(ll)y.s*x.t;
26 }
27 priority_queue<Node,vector<Node>,greater<Node> >q;
28 int n;
29 int par[N],fa[N],siz[N],sum[N],val[N*2],w[N],r[N];
30 int cnt,hed[N],nxt[N*2],rea[N*2];
31 ll res,ans[N];
32 
33 void add(int u,int v)
34 {
35     nxt[++cnt]=hed[u];
36     hed[u]=cnt;
37     rea[cnt]=v;
38 }
39 void dfs(int u,int f)
40 {
41     for (int i=hed[u];i!=-1;i=nxt[i])
42     {
43         int v=rea[i];
44         if (v==f)continue;
45         par[v]=u,dfs(v,u);
46     }
47 }
48 int find(int num)
49 {
50     if (fa[num]!=num)fa[num]=find(fa[num]);
51     return fa[num];
52 }
53 void solve(int rt)
54 {
55     for (int i=1;i<=n;i++)
56     {
57         fa[i]=i,siz[i]=1;
58         sum[i]=ans[i]=w[i];
59         q.push((Node){i,w[i],1});
60     }
61     dfs(rt,0);
62     for (int i=1;i<n;)
63     {
64         Node now=q.top();q.pop();
65         if (now.t!=siz[now.u]||now.u==rt)continue;
66         i++;
67         int u=find(par[now.u]),v=now.u;
68         ans[u]+=ans[v]+(ll)sum[v]*siz[u];
69         res=max(res,ans[u]);
70         fa[v]=u,siz[u]+=siz[v],sum[u]+=sum[v];
71         q.push((Node){u,sum[u],siz[u]});
72     }
73     while(!q.empty())q.pop();
74 }
75 int main()
76 {
77     memset(hed,-1,sizeof(hed));
78     n=read();
79     for (int i=1;i<n;i++)
80     {
81         int x=read(),y=read();
82         add(x,y),add(y,x);
83     }
84     for (int i=1;i<=n;i++)
85         w[i]=read(),r[i]=read();
86     for (int i=1;i<=n;i++)
87         if (r[i])solve(i);
88     printf("%lld\n",res);
89 }