sb\(O(n^{2})\)传参

T1

暴力一会儿就码好，结果..

祭奠一下死去的代码

died

#include<cstdio>
#define MAX 1010
#define re register
namespace OMA
{
   int n,q;
   int r,c,l,s;
   struct martix
   {
     int ar[MAX][MAX];
     inline friend martix operator +(martix a,int add)
     {
       for(re int i=r; i<=r+l-1; i++)
       {
         for(re int j=c; j<=i-r+c; j++)
         { a.ar[i][j] += add; }
       }
       return a;
     }
     inline int ans()
     {
       int sum = 0;
       for(re int i=1; i<=n; i++)
       {
         for(re int j=1; j<=n; j++)
         { sum ^= ar[i][j]; }
       }
       return sum;
     }
     inline void print()
     {
       for(re int i=1; i<=n; i++)
       {
         for(re int j=1; j<=n; j++)
         {
           printf("%d ",ar[i][j]);
         }
         printf("\n");
       }
     }
   }u;
   inline int read()
   {
     int s=0,w=1; char ch=getchar();
     while(ch<'0'||ch>'9'){ if(ch=='-')w=-1; ch=getchar(); }
     while(ch>='0'&&ch<='9'){ s=s*10+ch-'0'; ch=getchar(); }
     return s*w;
   }
   signed main()
   {
     n = read(),q = read();
     for(re int i=1; i<=q; i++)
     { r = read(),c = read(),l = read(),s = read(); u = u+s; }
     //u.print();
     printf("%d\n",u.ans());
     return 0;
   }
}
signed main()
{ return OMA::main(); } 

好吧，其实是我传参的时候传了个1000*1000的矩阵进去，然后就炸了，我真傻，真的打暴力还整花活，还打挂了，给1pts都算便宜我了。

正解：

差分。

Code

#include<cstdio>
#define MAX 2010
#define re register
#define int long long
namespace OMA
{
   int n,q,ans;
   int r,c,l,s;
   int sum[MAX][MAX][2];
   inline int read()
   {
     int s=0,w=1; char ch=getchar();
     while(ch<'0'||ch>'9'){ if(ch=='-')w=-1; ch=getchar(); }
     while(ch>='0'&&ch<='9'){ s=s*10+ch-'0'; ch=getchar(); }
     return s*w;
   }
   signed main()
   {
     n = read(),q = read();
     for(re int i=1; i<=q; i++)
     {
       r = read(),c = read(),l = read(),s = read();
       sum[r][c][0] += s,sum[r+l][c][0] -= s;
       sum[r][c+1][1] -= s,sum[r+l][c+l+1][1] += s;
     }
     for(re int i=1; i<=n; i++)
     {
       int tmp = 0;
       for(re int j=1; j<=n; j++)
       {
         sum[i][j][0] += sum[i-1][j][0],sum[i][j][1] += sum[i-1][j-1][1];
         ans ^= tmp += (sum[i][j][0]+sum[i][j][1]);
       }
     }
     printf("%lld\n",ans);
     return 0;
   }
}
signed main()
{ return OMA::main(); } 

T2

考场一看期望直接跳

明显的状压，但更明显的是，数组开不下，所以考虑map，但是会TLE几个点，所以考虑手写hash表，可是太慢了，所以考虑特判，很明显，对于 \(k=n\) 或 \(k=n-1\) 的点，答案即为白色球的个数。特判一下就好，跑得还飞快

特判前：

优化特判后：

Code

#include<cstdio>
#include<cstring>
#include<iostream>
#define re register
#define seed 1000007
namespace OMA
{
   struct Hash_map
   {
     struct node
     {
       int next;
       int len;
       int to;
       double v;
     }star[seed];
     int len,cnt,head[seed];
     inline double &operator [](int sta)
     {
       int key = 1LL*sta*len%seed;
       for(re int i=head[key]; i; i=star[i].next)
       {
         if(star[i].to==sta&&star[i].len==len)
         { return star[i].v; }
       }
       return star[++cnt] = (node){head[key],len,sta,-1},head[key] = cnt,star[cnt].v;
     }
   }map;
   int n,k;
   inline double dfs(int len,int sta)
   {
     if(len==n-k)
     { return 0; }
     map.len = len;
     if(map[sta]>-1.0)
     { return map[sta]; }
     int temp,tmp[50];
     map[temp = sta] = 0;
     for(re int i=1; i<=len; i++)
     { tmp[i] = temp&1,temp >>= 1; }
     for(re int i=1; i<=len/2; i++)
     {
       double tmp1 = dfs(len-1,((sta>>(len-i+1))<<(len-i))|(sta&((1<<len-i)-1)))+tmp[len-i+1];
       double tmp2 = dfs(len-1,((sta>>i)<<(i-1))|(sta&((1<<i-1)-1)))+tmp[i];
       map.len = len,map[sta] += 2.0/len*std::max(tmp1,tmp2);
     }
     if(len&1)
     {
       double tmp1 = dfs(len-1,(sta>>(len-((len>>1)+1)+1))<<(len-((len>>1)+1))|(sta&((1<<(len-((len>>1)+1)))-1)))+tmp[len/2+1];
       map.len = len,map[sta] += 1.0/len*tmp1;
     }
     return map[sta];
   }
   int sta,cnt;
   char ch[50];
   signed main()
   {
     scanf("%d%d%s",&n,&k,ch+1);
     for(re int i=1; i<=n; i++)
     { sta <<= 1; if(ch[i]=='W'){ cnt++,sta++; } }
     if(k&&(k==n||k==n-1))
     { printf("%0.10lf\n",(double)cnt); return 0; }
     printf("%0.10lf\n",dfs(n,sta));
     return 0;
   }
}
signed main()
{ return OMA::main(); } 

T3

考试的时候奔着1pts的点，结果加了个判断，0pts，你骗分还特判个什么啊

正解：

是个树形dp

快考试了，先咕了

Code

#include<cstdio>
#define MAX 100010
#define re register
#define INF 114514810
namespace OMA
{
   int n;
   struct Graph
   {
     int next;
     int to;
     int w;
   }edge[MAX<<1];
   int cnt=1,head[MAX];
   struct node
   {
     int a,b;
     friend bool operator <(const node &a,const node &b)
     { return a.a==b.a?a.b<b.b:a.a<b.a; }
   }dp[MAX][2];
   inline int read()
   {
     int s=0,w=1; char ch=getchar();
     while(ch<'0'||ch>'9'){ if(ch=='-')w=-1; ch=getchar(); }
     while(ch>='0'&&ch<='9'){ s=s*10+ch-'0'; ch=getchar(); }
     return s*w;
   }
   inline void add(int u,int v,int w)
   {
     edge[++cnt].next = head[u];
     edge[cnt].to = v;
     edge[cnt].w = w;
     head[u] = cnt;
   }
   inline node calc(node a,node b)
   { return (node){a.a+b.a,a.b+b.b}; }
   inline node min(node a,node b)
   { return a<b?a:b; }
   inline void dfs(int u,int fa,int w)
   {
     node a = (node){INF,INF},b = (node){0,0};
     for(re int i=head[u]; i; i=edge[i].next)
     {
       int v = edge[i].to,w = edge[i].w;
       if(v!=fa)
       {
         dfs(v,u,w);
         node tmp1 = min(calc(a,dp[v][0]),calc(b,dp[v][1]));
         node tmp2 = min(calc(b,dp[v][0]),calc(a,dp[v][1]));
         a = tmp1,b = tmp2;
       }
     }
     if(w==1)
     { dp[u][0] = (node){INF,INF}; }
     else
     { dp[u][0] = min(b,(node){a.a+1,a.b}); }
     if(!w)
     { dp[u][1] = (node){INF,INF}; }
     else
     { dp[u][1] = min((node){a.a,a.b+1},(node){b.a+1,b.b+1}); }
   }
   inline int w(int c,int d)
   { return d==2?d:c!=d; }
   signed main()
   {
     n = read();
     for(re int i=1; i<=n-1; i++)
     {
       int u = read(),v = read(),c = read(),d = read();
       add(u,v,w(c,d)),add(v,u,w(c,d));
     }
     dfs(1,0,2);
     printf("%d %d\n",dp[1][0].a>>1,dp[1][0].b);
     return 0;
   }
}
signed main()
{ return OMA::main(); }