• Educational Codeforces Round 71 (Rated for Div. 2)


    https://codeforces.com/contest/1207

    这次没打rating

    A、There Are Two Types Of Burgers

    数据规模不大,暴力即可

     1 #include<iostream>
     2 #include<sstream>
     3 #include<fstream>
     4 #include<algorithm>
     5 #include<cstring>
     6 #include<iomanip>
     7 #include<cstdlib>
     8 #include<cctype>
     9 #include<vector>
    10 #include<string>
    11 #include<cmath>
    12 #include<ctime>
    13 #include<stack>
    14 #include<queue>
    15 #include<map>
    16 #include<set>
    17 #define mem(a,b) memset(a,b,sizeof(a))
    18 #define random(a,b) (rand()%(b-a+1)+a)
    19 #define ll long long
    20 #define ull unsigned long long
    21 #define e 2.71828182
    22 #define Pi acos(-1.0)
    23 #define ls(rt) (rt<<1)
    24 #define rs(rt) (rt<<1|1)
    25 #define lowbit(x) (x&(-x))
    26 using namespace std;
    27 int read()
    28 {
    29     int s=1,x=0;
    30     char ch=getchar();
    31     while(!isdigit(ch)) {if(ch=='-') s=-1;ch=getchar();}
    32     while(isdigit(ch)) {x=10*x+ch-'0';ch=getchar();}
    33     return x*s;
    34 }
    35 int b,p,f,h,c;
    36 void solve()
    37 {
    38     int res=-1;
    39     for(int i=0;i<=min(p,b>>1);++i)
    40     {
    41         int j=min(f,(b-i*2)>>1);
    42         res=max(res,h*i+c*j);
    43     }
    44     cout<<res<<endl;
    45 }
    46 int main()
    47 {
    48     int test=read();
    49     while(test--)
    50     {
    51         b=read(),p=read(),f=read(),h=read(),c=read();
    52         solve();
    53     }
    54 }
    View Code

    B、Square Filling

     给定一个01矩阵a,全0矩阵b,每次操作能将2*2的元素变为1,求操作序列使得矩阵b变为a,若不能则输出-1

    刚好在做DLX的习题,想到转化为用DLX解重复覆盖问题。

      1 #include<iostream>
      2 #include<sstream>
      3 #include<fstream>
      4 #include<algorithm>
      5 #include<cstring>
      6 #include<iomanip>
      7 #include<cstdlib>
      8 #include<cctype>
      9 #include<vector>
     10 #include<string>
     11 #include<cmath>
     12 #include<ctime>
     13 #include<stack>
     14 #include<queue>
     15 #include<map>
     16 #include<set>
     17 #define mem(a,b) memset(a,b,sizeof(a))
     18 #define random(a,b) (rand()%(b-a+1)+a)
     19 #define ll long long
     20 #define ull unsigned long long
     21 #define e 2.71828182
     22 #define Pi acos(-1.0)
     23 #define ls(rt) (rt<<1)
     24 #define rs(rt) (rt<<1|1)
     25 #define lowbit(x) (x&(-x))
     26 using namespace std;
     27 const int MAXN=2555;
     28 const int MAXM=2555;
     29 const int MAX=7e6+5;
     30 const int INF=0x3f3f3f3f;
     31 int read()
     32 {
     33     int s=1,x=0;
     34     char ch=getchar();
     35     while(!isdigit(ch)) {if(ch=='-') s=-1;ch=getchar();}
     36     while(isdigit(ch)) {x=10*x+ch-'0';ch=getchar();}
     37     return x*s;
     38 }
     39 int N,M;
     40 struct DLX
     41 {
     42     int n,m,cnt,ansd;
     43     int U[MAX],D[MAX],R[MAX],L[MAX],row[MAX],col[MAX];
     44     int H[MAXN],S[MAXM],ans[MAXN],tmp[MAXN];
     45     bool v[MAX];
     46     void init(int _n,int _m)
     47     {
     48         n=_n,m=_m,ansd=INF;
     49         for(int i=0;i<=m;++i)
     50         S[i]=0,U[i]=D[i]=i,L[i]=i-1,R[i]=i+1;
     51         R[m]=0,L[0]=m,cnt=m;
     52         mem(H,-1);
     53     }
     54     void add(int r,int c)
     55     {
     56         ++S[col[++cnt]=c];
     57         row[cnt]=r;
     58         D[cnt]=D[c],U[D[c]]=cnt;
     59         U[cnt]=c,D[c]=cnt;
     60         if(H[r]<0) H[r]=L[cnt]=R[cnt]=cnt;
     61         else R[cnt]=R[H[r]],L[R[H[r]]]=cnt,L[cnt]=H[r],R[H[r]]=cnt;
     62     }
     63     void remove(int c)
     64     {
     65         for(int i=D[c];i!=c;i=D[i])
     66         L[R[i]]=L[i],R[L[i]]=R[i];
     67     }
     68     void resume(int c)
     69     {
     70         for(int i=U[c];i!=c;i=U[i])
     71         L[R[i]]=R[L[i]]=i;
     72     }
     73     int h()
     74     {
     75         int ret=0;
     76         for(int i=R[0];i!=0;i=R[i]) v[i]=true;
     77         for(int i=R[0];i!=0;i=R[i])
     78         {
     79             if(!v[i]) continue;
     80             ret++,v[i]=false;
     81             for(int j=D[i];j!=i;j=D[j])
     82             for(int k=R[j];k!=j;k=R[k])
     83             v[col[k]]=false;
     84         }    
     85         return ret;
     86     } 
     87     bool dance(int d)
     88     {
     89         //if(d+h()>=ansd) return;
     90         if(R[0]==0)
     91         {
     92             if(d<ansd) 
     93             {
     94                 ansd=d;
     95                 for(int i=0;i<d;++i) ans[i]=tmp[i];
     96             }
     97             return true;
     98         }
     99         int c=R[0];
    100         for(int i=R[0];i!=0;i=R[i])
    101         if(S[i]<S[c]) c=i;
    102         for(int i=D[c];i!=c;i=D[i])
    103         {
    104             remove(i);
    105             for(int j=R[i];j!=i;j=R[j]) remove(j);
    106             tmp[d]=row[i];
    107             if(dance(d+1)) return true;
    108             for(int j=L[i];j!=i;j=L[j]) resume(j);
    109             resume(i);
    110         }
    111         return false;
    112     }
    113 }dlx;
    114 int G[51][51];
    115 int pos[51][51];
    116 struct node
    117 {
    118     int x,y;
    119 }data[MAXN];
    120 int main()
    121 {
    122     N=read(),M=read();
    123     int cnt=0;
    124     for(int i=1;i<=N;++i)
    125     for(int j=1;j<=M;++j)
    126     {
    127         G[i][j]=read();
    128         if(G[i][j]) pos[i][j]=++cnt;
    129     }
    130     dlx.init(N*M,cnt);
    131     cnt=0;
    132     for(int i=1;i<=N-1;++i)
    133     for(int j=1;j<=M-1;++j)
    134     {
    135         bool flag=true;
    136         for(int x=1;x<=2&&flag;++x)
    137         for(int y=1;y<=2&&flag;++y)
    138         if(!G[i+x-1][j+y-1]) flag=false;
    139         
    140         if(flag)
    141         {
    142             dlx.add(++cnt,pos[i][j]);
    143             dlx.add(cnt,pos[i][j+1]);
    144             dlx.add(cnt,pos[i+1][j]);
    145             dlx.add(cnt,pos[i+1][j+1]);
    146             data[cnt].x=i,data[cnt].y=j;
    147         }
    148         
    149     }
    150     
    151     dlx.dance(0);
    152     if(dlx.ansd==INF) return 0*printf("%d",-1);
    153     cout<<dlx.ansd<<endl;
    154     for(int i=0;i<dlx.ansd;++i) 
    155     {
    156         int r=dlx.ans[i];
    157         cout<<data[r].x<<' '<<data[r].y<<endl;
    158     }
    159     
    160 }
    View Code

    其实只要模拟就行了,只有当矩阵a有2*2的元素全1时才在矩阵b对应位置改,比较最后的矩阵a和矩阵b,贴下大佬代码

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 #define ll long long
     4 #define pb push_back
     5 #define mp make_pair
     6 const int N=100;
     7 int a[N][N],b[N][N];
     8 int main()
     9 {
    10     int n,m;
    11     scanf("%i %i",&n,&m);
    12     for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) scanf("%i",&a[i][j]);
    13     vector<pair<int,int>> ans;
    14     for(int i=1;i<=n;i++) for(int j=1;j<=m;j++)
    15     {
    16         if(a[i][j]==1 && a[i+1][j]==1 && a[i][j+1]==1 && a[i+1][j+1]==1)
    17         {
    18             b[i][j]=b[i+1][j]=b[i][j+1]=b[i+1][j+1]=1;
    19             ans.pb({i,j});
    20         }
    21     }
    22     for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) if(a[i][j]!=b[i][j]) return 0*printf("-1
    ");
    23     printf("%i
    ",ans.size());
    24     for(auto p:ans) printf("%i %i
    ",p.first,p.second);
    25     return 0;
    26 }
    View Code

    C、Gas Pipeline

    给定一个长为n的01串,emm,看图吧

    令str表示01串,则str[i]=1表示第i个区间高度必须为2,str[i]=0表示第i个区间高度可以不必为2。(i>=1)

    用dp[i][1]表示第i个区间后半段高度为1时的最小花费,dp[i][2]表示第i个区间后半段高度为2时的最小花费。

    题目中给定第1个区间左端柱子高度和第n个区间右端柱子高度是1,因此初始化dp[0][1]=b,dp[0][2]=INF,最后的结果即是dp[n][1]。

    递推式列出情况推一推就好:

     1 #include<iostream>
     2 #include<sstream>
     3 #include<fstream>
     4 #include<algorithm>
     5 #include<cstring>
     6 #include<iomanip>
     7 #include<cstdlib>
     8 #include<cctype>
     9 #include<vector>
    10 #include<string>
    11 #include<cmath>
    12 #include<ctime>
    13 #include<stack>
    14 #include<queue>
    15 #include<map>
    16 #include<set>
    17 #define mem(a,b) memset(a,b,sizeof(a))
    18 #define random(a,b) (rand()%(b-a+1)+a)
    19 #define ll long long
    20 #define ull unsigned long long
    21 #define e 2.71828182
    22 #define Pi acos(-1.0)
    23 #define ls(rt) (rt<<1)
    24 #define rs(rt) (rt<<1|1)
    25 #define lowbit(x) (x&(-x))
    26 using namespace std;
    27 const int MAXN=2e5+5;
    28 const ll INF=1ll<<62;
    29 char str[MAXN];
    30 ll dp[MAXN][3];
    31 ll read()
    32 {
    33     ll s=1,x=0;
    34     char ch=getchar();
    35     while(!isdigit(ch)) {if(ch=='-') s=-1;ch=getchar();}
    36     while(isdigit(ch)) {x=10*x+ch-'0';ch=getchar();}
    37     return x*s;
    38 }
    39 int main()
    40 {
    41     ll test=read();
    42     while(test--)
    43     {
    44         ll n=read(),a=read(),b=read();
    45         scanf("%s",str+1);    
    46         dp[0][1]=b,dp[0][2]=INF,str[0]='0';
    47         for(int i=1;i<=n;++i)
    48         {
    49             if(str[i]=='1')
    50             {
    51                 dp[i][1]=INF;            
    52                 dp[i][2]=dp[i-1][2]+a+2*b;
    53             }
    54             else if(str[i]=='0')
    55             {
    56                 if(str[i-1]=='0')
    57                 {   
    58                     dp[i][1]=min(dp[i-1][1]+a+b,dp[i-1][2]+2*a+b);
    59                     dp[i][2]=min(dp[i-1][1]+2*a+2*b,dp[i-1][2]+a+2*b);
    60                 }
    61                 else
    62                 {
    63                     dp[i][1]=dp[i-1][2]+2*a+b;
    64                     dp[i][2]=dp[i-1][2]+a+2*b;
    65                 }
    66             }
    67         }
    68         /*for(int i=1;i<=n;++i)
    69         cout<<dp[i][1]<<' ';cout<<endl;
    70         for(int i=1;i<=n;++i)
    71         cout<<dp[i][2]<<' ';cout<<endl;*/
    72         cout<<dp[n][1]<<endl;
    73     } 
    74 }
    View Code

    D、Number Of Permutations

    定义一个pair序列为bad:pair的first元素为非降,或second元素为非降

    给定一个pair序列,求使得该序列为good(非bad)的排列数,模998244353

    正难则反

    容斥原理

    记一个RE的bug:

     1 #include<iostream>
     2 #include<sstream>
     3 #include<fstream>
     4 #include<algorithm>
     5 #include<cstring>
     6 #include<iomanip>
     7 #include<cstdlib>
     8 #include<cctype>
     9 #include<vector>
    10 #include<string>
    11 #include<cmath>
    12 #include<ctime>
    13 #include<stack>
    14 #include<queue>
    15 #include<map>
    16 #include<set>
    17 #define mem(a,b) memset(a,b,sizeof(a))
    18 #define random(a,b) (rand()%(b-a+1)+a)
    19 #define ll long long
    20 #define ull unsigned long long
    21 #define e 2.71828182
    22 #define Pi acos(-1.0)
    23 #define ls(rt) (rt<<1)
    24 #define rs(rt) (rt<<1|1)
    25 #define P pair<int,int>
    26 #define lowbit(x) (x&(-x))
    27 using namespace std;
    28 const int MAXN=1e6+5; 
    29 const int MOD=998244353;
    30 P seq[MAXN];
    31 ll fac[MAXN];
    32 int n;
    33 int read()
    34 {
    35     int s=1,x=0;
    36     char ch=getchar();
    37     while(!isdigit(ch)) {if(ch=='-') s=-1;ch=getchar();}
    38     while(isdigit(ch)) {x=10*x+ch-'0';ch=getchar();}
    39     return x*s;
    40 }
    41 void init()
    42 {
    43     fac[0]=1;
    44     for(int i=1;i<=n;++i)
    45     fac[i]=fac[i-1]*i%MOD;
    46 }
    47 bool cmp(P a,P b)
    48 {
    49     return a.second<b.second;
    50 }
    51 int main()
    52 {
    53     n=read();
    54     for(int i=1;i<=n;++i)
    55     seq[i].first=read(),seq[i].second=read();
    56     init();
    57     //calculate cnt2
    58     ll cnt2=1,tmp=1;
    59     sort(seq+1,seq+n+1,cmp);
    60     for(int i=2;i<=n;++i)
    61     if(seq[i].second==seq[i-1].second) tmp++;
    62     else cnt2=cnt2*fac[tmp]%MOD,tmp=1;
    63     cnt2=cnt2*fac[tmp]%MOD;
    64     
    65     //calculate cnt1
    66     ll cnt1=1;tmp=1;
    67     sort(seq+1,seq+1+n);
    68     for(int i=2;i<=n;++i)
    69     if(seq[i].first==seq[i-1].first) tmp++;
    70     else cnt1=cnt1*fac[tmp]%MOD,tmp=1;
    71     cnt1=cnt1*fac[tmp]%MOD;
    72         
    73     //calculate cnt3
    74     ll cnt12=1;tmp=1;
    75     bool flag=true;
    76     for(int i=2;i<=n&&flag;++i)
    77     {
    78         if(seq[i].second<seq[i-1].second) flag=false;
    79         else if(seq[i].first==seq[i-1].first&&
    80                 seq[i].second==seq[i-1].second)
    81             tmp++;
    82         else cnt12=cnt12*fac[tmp]%MOD,tmp=1;
    83     }
    84     cnt12=cnt12*fac[tmp]%MOD;
    85     if(!flag) cnt12=0;
    86     
    87     cout<<(fac[n]-cnt1-cnt2+cnt12+MOD*2)%MOD<<endl;
    88     return 0;
    89 }
    View Code
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  • 原文地址:https://www.cnblogs.com/wangzhebufangqi/p/11402694.html
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