NOIP模拟赛-2018.11.5

NOIP模拟赛

　　好像最近每天都会有模拟赛了.今天从高二逃考试跑到高一机房，然而高一也要考试，这回好像没有拒绝的理由了。

　　今天的模拟赛好像很有技术含量的感觉.

　　T1:xgy断句.

　　好诡异的题目,首先给出一些词,一个字符串,要求断句:每个句子至少有三个词,词数是总单词数的因数,单词得是字典里的词.求最多能断多少句.

　　首先当然是暴力匹配每一段是否是单词,然后$f_i$表示以$i$结尾的前缀中最多能断多少句,枚举断点进行转移,如何判断能否构成句子呢?搜索啊.

　　这里一定要注意如果最后一个状态是极小值,那么输出$0$.因为这个挂了$20$分.

　　

 # include <cstdio>
 # include <iostream>
 # include <cstring>
 # include <string>
 # include <algorithm>
 # include <cmath>
 # define R register int
 # define ll long long

 using namespace std;

 const int maxn=;
 int n,len[maxn],l;
 char dic[maxn][];
 char s[];
 int dp[],leng,checkf;
 int f[][],c[][];

 void dfs (int x,int s,int r)
 {
     if(x==r+)
     {
         if(s>=&&leng%s==) checkf=;
         return ;
     }
     if(checkf) return;
     for (R i=x;i<=r;++i)
         if(c[x][i])
             dfs(i+,s+,r);
 }

 bool check (int l,int r)
 {
     if(f[l][r]!=-) return f[l][r];
     leng=r-l+;
     checkf=false;
     dfs(l,,r);
     f[l][r]=checkf;
     return f[l][r];
 }

 void init()
 {
     for (R i=;i<=l;++i)
         for (R j=;j<=i;++j)
         {
             int f=false;
             for (R k=;k<=n;++k)
             {
                 if(len[k]!=i-j+) continue;
                 int ff=true;
                 for (R z=;z<=len[k];++z)
                     if(s[j+z-]!=dic[k][z]) ff=false;
                 if(ff) f=true;
                 if(f) break;
             }
             if(f) c[j][i]=;
         }
 }

 int main()
 {
     freopen("xgy.in","r",stdin);
     freopen("xgy.out","w",stdout);

     scanf("%d",&n);
     memset(f,-,sizeof(f));
     for (R i=;i<=n;++i)
     {
         scanf("%s",dic[i]+);
         len[i]=strlen(dic[i]+);
         for (R j=;j<=len[i];++j)
             if(dic[i][j]>='A'&&dic[i][j]<='Z') dic[i][j]=dic[i][j]-'A'+'a';
     }
     scanf("%s",s+);
     l=strlen(s+);
     for (R i=;i<=l;++i)
         if(s[i]>='A'&&s[i]<='Z') s[i]=s[i]-'A'+'a';
     init();
     for (R i=;i<=l;++i)
         dp[i]=-;
     dp[]=;
     for (R i=;i<=l;++i)
         for (R j=;j<i;++j)
         {
             if(dp[j]==-) continue;
             if(i-j+>) continue;
             if(check(j+,i)) dp[i]=max(dp[i],dp[j]+);
         }
     printf("%d\n",dp[l]);
     fclose(stdin);
     fclose(stdout);
     return ;
 }

xgy断句

　　T2:多人背包.

　　显然前$k$优的状态不可能由比前$k$个还劣的状态转移过来,那么只保存前$k$优的即可.然而我强行把$KlogK$写成了$K^3$,竟然卡过了.

　　

 # include <cstdio>
 # include <iostream>
 # include <cstring>
 # include <string>
 # include <algorithm>
 # include <cmath>
 # define R register int
 # define ll long long

 using namespace std;

 const int maxn=;
 int k,V,n;
 int w[maxn],v[maxn];
 int dp[][];
 int a[maxn];

 int main()
 {
     freopen("bag.in","r",stdin);
     freopen("bag.out","w",stdout);

     scanf("%d%d%d",&k,&V,&n);
     for (R i=;i<=n;++i) scanf("%d%d",&w[i],&v[i]);
     memset(dp,,sizeof(dp));
     dp[][]=;
     for (R i=;i<=n;++i)
         for (R j=V;j>=w[i];--j)
         {
             for (R z=;z<=k;++z)
             {
                 int T=dp[z][ j-w[i] ]+v[i];
                 if(T<) continue;
                 for (R l=z;l<=k;++l)
                     if(T>dp[l][j])
                     {
                         for (R t=k;t>l;--t)
                             dp[t][j]=dp[t-][j];
                         dp[l][j]=T;
                         break;
                     }
             }
         }
     int ans=;
     for (R i=;i<=k;++i) ans+=dp[i][V];
     printf("%d",ans);
     fclose(stdin);
     fclose(stdout);
     return ;
 }

多人背包

　　T3:冰原探险.

　　似乎爆搜加$map$判重可过,离散化比较麻烦,而且把一些山之间的通道给离散没了...得了70.

　　

 # include <cstdio>
 # include <iostream>
 # include <map>
 # include <algorithm>
 # include <bitset>
 # include <queue>
 # define R register int

 using namespace std;

 const int dx[]={-,,,};
 const int dy[]={,-,,};
 const int maxn=;
 const int knc=;
 int n,N,M;
 map <int,int> m;
 int x[maxn],y[maxn],sx,sy,ex,ey,a[maxn],b[maxn],c[maxn],d[maxn];
 bool ic[knc][knc],vis[knc][knc];
 struct node
 {
     int x,y,v;
 };
 queue <node> q;

 int bfs ()
 {
     node a,b;
     a.x=sx,a.y=sy,a.v=;
     q.push(a);
     while(q.size())
     {
         a=q.front();
         q.pop();
         int x=a.x,y=a.y;
         vis[x][y]=;
         if(a.x==ex&&a.y==ey) return a.v;
         for (R d=;d<;++d)
         {
             x=a.x,y=a.y;
             while(ic[x][y]==)
             {
                 x+=dx[d],y+=dy[d];
                 if(x>N||y>M||x<||y<) break;
                 if(x==ex&&y==ey)
                     return a.v+;
             }
             if(x>N||y>M||x<||y<) continue;
             if(!ic[x][y]) continue;
             x-=dx[d],y-=dy[d];
             if(vis[x][y]) continue;
             vis[x][y]=;
             b.x=x,b.y=y,b.v=a.v+;
             q.push(b);
         }
     }
     return ;
 }

 int read ()
 {
     R x=,f=;
     char c=getchar();
     while(!isdigit(c)) { if(c=='-') f=-f; c=getchar(); }
     while(isdigit(c)) x=(x<<)+(x<<)+(c^),c=getchar();
     return x*f;
 }

 int main()
 {
     freopen("ice.in","r",stdin);
     freopen("ice.out","w",stdout);

     n=read();
     sx=read(),sy=read(),ex=read(),ey=read();
     x[++x[]]=sx,x[++x[]]=ex;
     y[++y[]]=sy,y[++y[]]=ey;
     for (R i=;i<=n;++i)
     {
         a[i]=read(),b[i]=read(),c[i]=read(),d[i]=read();
         x[++x[]]=a[i],x[++x[]]=c[i];
         y[++y[]]=b[i],y[++y[]]=d[i];
     }
     sort(x+,x++x[]);
     sort(y+,y++y[]);
     int cnt=;
     for (R i=;i<=x[];++i)
     {
         if(x[i]!=x[i-]||i==) cnt++;
         m[ x[i] ]=cnt;
     }
     N=cnt;
     for (R i=;i<=n;++i)
         a[i]=m[ a[i] ],c[i]=m[ c[i] ];
     sx=m[sx],ex=m[ex];
     cnt=;
     m.clear();
     for (R i=;i<=y[];++i)
     {
         if(y[i]!=y[i-]||i==) cnt++;
         m[ y[i] ]=cnt;
     }
     M=cnt;
     for (R i=;i<=n;++i)
         b[i]=m[ b[i] ],d[i]=m[ d[i] ];
     sy=m[sy],ey=m[ey];
     for (R i=;i<=n;++i)
         for (R j=a[i];j<=c[i];++j)
             for (R z=b[i];z<=d[i];++z)
                 ic[j][z]=;

     printf("%d",bfs());
     fclose(stdin);
     fclose(stdout);
     return ;
 }

冰原探险

---shzr