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  • Codeforces Beta Round #3

    Codeforces Beta Round #3

    http://codeforces.com/contest/3

    A

    找规律题。但我懒得找,直接暴搜

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 #define lson l,mid,rt<<1
     4 #define rson mid+1,r,rt<<1|1
     5 #define sqr(x) ((x)*(x))
     6 typedef long long ll;
     7 /*#ifndef ONLINE_JUDGE
     8         freopen("1.txt","r",stdin);
     9 #endif */
    10 
    11 int book[15][15];
    12 struct sair{
    13     string str[105];
    14     int co;
    15     int x,y;
    16 };
    17 
    18 int dir[8][2]={0,1,1,0,0,-1,-1,0,1,1,-1,-1,1,-1,-1,1};
    19 string D[8]={"R","D","L","U","RD","LU","LD","RU"};
    20 
    21 sair bfs(sair s){
    22     queue<sair>Q;
    23     Q.push(s);
    24     sair e;
    25     while(!Q.empty()){
    26         s=Q.front();
    27         Q.pop();
    28         for(int i=0;i<8;i++){
    29             e=s;
    30             e.x=s.x+dir[i][0];
    31             e.y=s.y+dir[i][1];
    32             if(e.x>=0&&e.x<8&&e.y>=0&&e.y<8&&book[e.x][e.y]!=1){
    33                 e.str[e.co]=D[i];
    34                 e.co++;
    35                 Q.push(e);
    36                 if(book[e.x][e.y]==2){
    37                     return e;
    38                 }
    39                 book[e.x][e.y]=1;
    40             }
    41         }
    42     }
    43 }
    44 
    45 int main(){
    46     #ifndef ONLINE_JUDGE
    47         freopen("1.txt","r",stdin);
    48     #endif
    49     string str,str1;
    50     cin>>str;
    51     sair s;
    52     book[7-(str[1]-'1')][str[0]-'a']=1;
    53     s.y=str[0]-'a',s.x=7-(str[1]-'1'),s.co=0;
    54     cin>>str1;
    55     book[7-(str1[1]-'1')][str1[0]-'a']=2;
    56     if(str==str1) {cout<<0<<endl;return 0;}
    57     s=bfs(s);
    58     cout<<s.co<<endl;
    59     for(int i=0;i<s.co;i++){
    60         cout<<s.str[i]<<endl;
    61     }
    62 }
    View Code

    B

    贪心题

    因为体积只有1和2两种,所以每次选单位价值最大的物品。需要注意最后容量剩一个的情况,这时候需要对比是选一个好还是把容量为1的物品删去一个再添加一个容量为2的物品好

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 #define lson l,mid,rt<<1
     4 #define rson mid+1,r,rt<<1|1
     5 #define sqr(x) ((x)*(x))
     6 typedef long long ll;
     7 /*#ifndef ONLINE_JUDGE
     8         freopen("1.txt","r",stdin);
     9 #endif */
    10 
    11 string str;
    12 vector<pair<ll,int> >one,two;
    13 
    14 bool cmp(pair<ll,int>a,pair<ll,int>b){
    15     if(a.first==b.first) return a.second<b.second;
    16     return a.first>b.first;
    17 }
    18 
    19 int main(){
    20     #ifndef ONLINE_JUDGE
    21         freopen("1.txt","r",stdin);
    22     #endif
    23     ll n,m;
    24     cin>>n>>m;
    25     int v;
    26     ll val;
    27     for(int i=1;i<=n;i++){
    28         cin>>v>>val;
    29         if(v==1){
    30             one.push_back(make_pair(val,i));
    31         }
    32         else{
    33             two.push_back(make_pair(val,i));
    34         }
    35     }
    36     sort(one.begin(),one.end(),cmp);
    37     sort(two.begin(),two.end(),cmp);
    38     int pos1=0,pos2=0;
    39     ll ans=0;
    40     while(pos1<one.size()||pos2<two.size()){
    41       //  cout<<pos1<<" "<<pos2<<endl;
    42         if(m>=2){
    43             if((two[pos2].first/2)>=one[pos1].first&&pos2<two.size()){
    44                 ans+=two[pos2].first;
    45                 pos2++;
    46                 m-=2;
    47             }
    48             else if(one.size()>pos1){
    49                 ans+=one[pos1].first;
    50                 pos1++;
    51                 m-=1;
    52             }
    53             else{
    54                 ans+=two[pos2].first;
    55                 pos2++;
    56                 m-=2;
    57             }
    58         }
    59         else if(m==1){
    60             if(pos2==two.size()){ans+=one[pos1].first;pos1++;}
    61             else if(pos1==one.size()){
    62                 if(pos1==0)
    63                     break;
    64                 else{
    65                     pos1--;
    66                     ans-=one[pos1].first;
    67                     ans+=two[pos2].first;
    68                     pos2++;
    69                 }
    70             }
    71             else if(two[pos2].first>one[pos1-1].first+one[pos1].first){
    72                 ans-=one[pos1-1].first;
    73                 pos1--;
    74                 ans+=two[pos2].first;
    75                 pos2++;
    76             }
    77             else{
    78                 ans+=one[pos1].first;
    79                 pos1++;
    80             }
    81             m--;
    82         }
    83         else if(m==0) break;
    84     }
    85     cout<<ans<<endl;
    86     vector<int>pos;
    87     for(int i=0;i<pos1;i++) pos.push_back(one[i].second);
    88     for(int i=0;i<pos2;i++) pos.push_back(two[i].second);
    89     sort(pos.begin(),pos.end());
    90     for(int i=0;i<pos.size();i++){
    91         cout<<pos[i]<<" ";
    92     }
    93     cout<<endl;
    94 }
    View Code

    C

    井子棋

    模拟判断各种情况就好

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 #define lson l,mid,rt<<1
     4 #define rson mid+1,r,rt<<1|1
     5 #define sqr(x) ((x)*(x))
     6 typedef long long ll;
     7 /*#ifndef ONLINE_JUDGE
     8         freopen("1.txt","r",stdin);
     9 #endif */
    10 
    11 string str[3];
    12 
    13 int Check(){
    14     int cha=0,quan=0;
    15     if(str[0][0]==str[0][1]&&str[0][1]==str[0][2]&&str[0][0]=='X') cha=1;
    16     if(str[1][0]==str[1][1]&&str[1][1]==str[1][2]&&str[1][0]=='X') cha=1;
    17     if(str[2][0]==str[2][1]&&str[2][1]==str[2][2]&&str[2][0]=='X') cha=1;
    18     if(str[0][0]==str[1][1]&&str[1][1]==str[2][2]&&str[0][0]=='X') cha=1;
    19     if(str[0][0]==str[1][0]&&str[1][0]==str[2][0]&&str[0][0]=='X') cha=1;
    20     if(str[0][1]==str[1][1]&&str[1][1]==str[2][1]&&str[0][1]=='X') cha=1;
    21     if(str[0][2]==str[1][2]&&str[1][2]==str[2][2]&&str[0][2]=='X') cha=1;
    22     if(str[0][2]==str[1][1]&&str[1][1]==str[2][0]&&str[0][2]=='X') cha=1;
    23 
    24     if(str[0][0]==str[0][1]&&str[0][1]==str[0][2]&&str[0][0]=='0') quan=1;
    25     if(str[1][0]==str[1][1]&&str[1][1]==str[1][2]&&str[1][0]=='0') quan=1;
    26     if(str[2][0]==str[2][1]&&str[2][1]==str[2][2]&&str[2][0]=='0') quan=1;
    27     if(str[0][0]==str[1][1]&&str[1][1]==str[2][2]&&str[0][0]=='0') quan=1;
    28     if(str[0][0]==str[1][0]&&str[1][0]==str[2][0]&&str[0][0]=='0') quan=1;
    29     if(str[0][1]==str[1][1]&&str[1][1]==str[2][1]&&str[0][1]=='0') quan=1;
    30     if(str[0][2]==str[1][2]&&str[1][2]==str[2][2]&&str[0][2]=='0') quan=1;
    31     if(str[0][2]==str[1][1]&&str[1][1]==str[2][0]&&str[0][2]=='0') quan=1;
    32     if(quan&&cha) return -1;
    33     if(cha) return 1;
    34     if(quan) return 2;
    35     return 0;
    36 }
    37 
    38 int main(){
    39     #ifndef ONLINE_JUDGE
    40        // freopen("1.txt","r",stdin);
    41     #endif
    42     ///first second illegal the first player won the second player won  draw
    43     for(int i=0;i<3;i++) cin>>str[i];
    44     int cha=0,quan=0;
    45     for(int i=0;i<3;i++) for(int j=0;j<3;j++) if(str[i][j]=='X') cha++; else if(str[i][j]=='0') quan++;
    46     if(abs(quan-cha)>=2||quan>cha) cout<<"illegal"<<endl;
    47     else{
    48         int tmp=Check();
    49         if(tmp==-1) cout<<"illegal"<<endl;
    50         else if(tmp==1) if(cha>quan) cout<<"the first player won"<<endl;else cout<<"illegal"<<endl;
    51         else if(tmp==2) if(quan==cha) cout<<"the second player won"<<endl;else cout<<"illegal"<<endl;
    52         else{
    53             if(quan+cha==9){
    54                 cout<<"draw"<<endl;
    55             }
    56             else if(cha>quan) cout<<"second"<<endl;
    57             else if(cha==quan) cout<<"first"<<endl;
    58         }
    59     }
    60 }
    View Code

    D

    括号匹配

    个人感觉是道好题

    先把所有的“?”变成“)”,然后当num<0时,把前面的右括号变成左括号,变的原则是b-a的最大值,最后判断是否括号匹配

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 #define lson l,mid,rt<<1
     4 #define rson mid+1,r,rt<<1|1
     5 #define sqr(x) ((x)*(x))
     6 typedef long long ll;
     7 /*#ifndef ONLINE_JUDGE
     8         freopen("1.txt","r",stdin);
     9 #endif */
    10 struct sair{
    11     ll a,b;
    12     int pos;
    13     bool operator<(const sair&bb)const{
    14         return (b-a)<(bb.b-bb.a);
    15     }
    16 };
    17 
    18 int main(){
    19     #ifndef ONLINE_JUDGE
    20         freopen("1.txt","r",stdin);
    21     #endif
    22     string str;
    23     cin>>str;
    24     ll a,b,ans=0;
    25     priority_queue<sair>Q;
    26     sair tmp;
    27     int num=0;
    28     for(int i=0;i<str.length();i++){
    29         if(str[i]=='('){
    30             num++;
    31         }
    32         else if(str[i]==')'){
    33             num--;
    34         }
    35         else{
    36             cin>>a>>b;
    37             tmp.a=a,tmp.b=b,tmp.pos=i;
    38             Q.push(tmp);
    39             ans+=b;
    40             str[i]=')';
    41             num--;
    42         }
    43         while(num<0){
    44             if(Q.size()){
    45                 ans-=Q.top().b-Q.top().a;
    46                 str[Q.top().pos]='(';
    47                 Q.pop();
    48                 num+=2;
    49             }
    50             else{
    51                 num=-0x3f3f3f3f;
    52                 break;
    53             }
    54         }
    55     }
    56     if(num!=0) cout<<-1<<endl;
    57     else{
    58         cout<<ans<<endl;
    59         cout<<str<<endl;
    60     }
    61 }
    View Code
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  • 原文地址:https://www.cnblogs.com/Fighting-sh/p/10339466.html
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