2017 ACM/ICPC Asia Regional Qingdao Online

Posted 2020-10-08 Psong

A Apple

不会Java和高精度

B Bomberman

unsolved

C The Dominator of Strings

取最长的串跑AC自动机，跑过的点需要标记。

  1 #include <bits/stdc++.h>
  2 #define Next next
  3 using namespace std;
  4 const int MAXN=1e5+5;
  5 struct Trie
  6 {
  7     int vis[MAXN];
  8     int next[MAXN][26],fail[MAXN],end[MAXN];
  9     int root,L;
 10     int newnode()
 11     {
 12         for(int i = 0;i < 26;i++)
 13             next[L][i] = -1;
 14         vis[L] = 0;
 15         end[L++] = 0;
 16         return L-1;
 17     }
 18     void init()
 19     {
 20         L = 0;
 21         root = newnode();
 22     }
 23     void insert(char buf[],int len)
 24     {
 25         int now = root;
 26         for(int i = 0;i < len;i++)
 27         {
 28             if(next[now][buf[i]-‘a‘] == -1)
 29                 next[now][buf[i]-‘a‘] = newnode();
 30             now = next[now][buf[i]-‘a‘];
 31         }
 32         end[now]++;
 33     }
 34     void build()
 35     {
 36         queue<int>Q;
 37         fail[root] = root;
 38         for(int i = 0;i < 26;i++)
 39             if(next[root][i] == -1)
 40                 next[root][i] = root;
 41             else
 42             {
 43                 fail[next[root][i]] = root;
 44                 Q.push(next[root][i]);
 45             }
 46         while( !Q.empty() )
 47         {
 48             int now = Q.front();
 49             Q.pop();
 50             for(int i = 0;i < 26;i++)
 51                 if(next[now][i] == -1)
 52                     next[now][i] = next[fail[now]][i];
 53                 else
 54                 {
 55                     fail[next[now][i]]=next[fail[now]][i];
 56                     Q.push(next[now][i]);
 57                 }
 58         }
 59     }
 60     int query(char buf[],int len)
 61     {
 62         int now = root;
 63         int res = 0;
 64         for(int i = 0;i < len;i++)
 65         {
 66             now = next[now][buf[i]-‘a‘];
 67             int temp = now;
 68             while( temp != root )
 69             {
 70                 if(vis[temp]) break;
 71                 vis[temp] = 1;
 72                 res += end[temp];
 73                 end[temp] = 0;
 74                 temp = fail[temp];
 75             }
 76         }
 77         return res;
 78     }
 79 }aho;
 80 
 81 int n;
 82 char tmp[MAXN];
 83 char str[MAXN];
 84 int main(){
 85     int t;
 86     scanf("%d",&t);
 87     while(t--){
 88         int Len=0;
 89         aho.init();
 90         scanf("%d",&n);
 91         for(int i=0;i<n;i++){
 92             scanf("%s",str);
 93             int len=strlen(str);
 94             aho.insert(str,len);
 95             if(Len<=len){
 96                 Len=len;
 97                 strcpy(tmp,str);
 98             }
 99         }
100         aho.build();
101         int ans=aho.query(tmp,Len);
102         if(ans==n){
103             printf("%s\n",tmp);
104         }
105         else printf("No\n");
106     }
107 
108     return 0;
109 }

Psong

D The Intersection

unsolved

E King‘s Visit

unsolved

F Pythagoras

unsolved

G Zuma

区间dp，先处理出连续相同的块。

1.普通合并。

2.两侧相同的情况，合并两侧。

3.两侧和中间合并，且需要保证两侧和不超过3个并且中间的个数只能是1。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int inf=0x3f3f3f3f;
 5 
 6 int n;
 7 char s[205];
 8 
 9 int tot;
10 int a[205];
11 int num[205];
12 int dp[205][205];
13 
14 int main(){
15     int t;
16     scanf("%d",&t);
17     int T=t;
18     while(t--){
19         tot=0;
20         memset(num,0,sizeof(num));
21         memset(dp,inf,sizeof(dp));
22         scanf("%s",s+1);
23         n=strlen(s+1);
24         int cnt=1;
25         for(int i=2;i<=n;i++){
26             if(s[i]!=s[i-1]){
27                 cnt%=3;
28                 if(cnt==0) continue;
29                 tot++;
30                 a[tot]=s[i-1]-‘0‘;
31                 num[tot]=cnt;
32                 cnt=1;
33             }
34             else cnt++;
35             if(i==n){
36                 if(cnt>=3) continue;
37                 tot++;
38                 a[tot]=s[i]-‘0‘;
39                 num[tot]=cnt;
40             }
41         }
42         for(int i=1;i<=tot;i++){
43             if(num[i]==1) dp[i][i]=2;
44             else dp[i][i]=1;
45         }
46         for(int len=2;len<=tot;len++){
47             for(int i=1;i<=tot-len+1;i++){
48                 int j=i+len-1;
49                 for(int k=i;k<j;k++){
50                     dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);
51                 }
52                 if(a[i]!=a[j]) continue;
53                 dp[i][j]=min(dp[i][j],dp[i+1][j-1]+(num[i]+num[j]<3));
54                 if(num[i]+num[j]<4){
55                     for(int k=i+1;k<j;k++){
56                         if(num[k]==1&&a[k]==a[i]){
57                             dp[i][j]=min(dp[i][j],dp[i+1][k-1]+dp[k+1][j-1]);
58                         }
59                     }
60                 }
61             }
62         }
63         printf("Case #%d: %d\n",T-t,dp[1][tot]);
64     }
65 
66     return 0;
67 }

Psong

H Chinese Zodiac

直接模拟

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <map>
 6 using namespace std;
 7 map<string,int> mp;
 8 int main(){
 9     mp["rat"]=11;
10     mp["ox"]=10;
11     mp["tiger"]=9;
12     mp["rabbit"]=8;
13     mp["dragon"]=7;
14     mp["snake"]=6;
15     mp["horse"]=5;
16     mp["sheep"]=4;
17     mp["monkey"]=3;
18     mp["rooster"]=2;
19     mp["dog"]=1;
20     mp["pig"]=0;
21     int t;
22     cin>>t;
23     while(t--){
24         string s1,s2;
25         cin>>s1>>s2;
26         int ans=mp[s1]+12-mp[s2];
27         if(ans>12) ans-=12;
28         cout<<ans<<endl;
29     }
30 
31 
32     return 0;
33 }

Psong

I Smallest Minimum Cut

边权*1000+1，结果为最大流对1000取模。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int inf=0x3f3f3f3f;
 4 const int MAXN=205;
 5 int level[MAXN];
 6 int iter[MAXN];
 7 int head[MAXN],tot;
 8 struct edge{
 9     int to,cap,Next;
10 } e[2005];
11 void init(){
12     memset(head,-1,sizeof(head));
13     tot=0;
14 }
15 void add(int from,int to,int cap){
16     e[tot].Next=head[from];
17     e[tot].to=to;
18     e[tot].cap=cap;
19     head[from]=tot;
20     tot++;
21 }
22 void addedge(int from,int to,int cap){
23     add(from,to,cap);
24     add(to,from,0);
25 }
26 void bfs(int s){
27     memset(level,-1,sizeof(level));
28     queue<int> q;
29     level[s]=0;
30     q.push(s);
31     while(!q.empty()){
32         int v=q.front(); q.pop();
33         for(int i=head[v];~i;i=e[i].Next){
34             edge &ed=e[i];
35             if(ed.cap>0&&level[ed.to]<0){
36                 level[ed.to]=level[v]+1;
37                 q.push(ed.to);
38             }
39         }
40     }
41 }
42 int dfs(int v,int t,int f){
43     if(v==t) return f;
44     for(int &i=iter[v];~i;i=e[i].Next){
45         edge &ed=e[i];
46         if(ed.cap>0&&level[v]<level[ed.to]){
47             int d=dfs(ed.to,t,min(f,ed.cap));
48             if(d>0){
49                 ed.cap-=d;
50                 e[i^1].cap+=d;
51                 return d;
52             }
53         }
54     }
55     return 0;
56 }
57 int max_flow(int s,int t){
58     int flow=0;
59     while(1){
60         bfs(s);
61         if(level[t]<0) return flow;
62         memcpy(iter,head,sizeof(iter));
63         int f;
64         while((f=dfs(s,t,inf))>0){
65             flow+=f;
66         }
67     }
68 }
69 
70 int n,m;
71 
72 int main(){
73     int t;
74     scanf("%d",&t);
75     while(t--){
76         init();
77         int s,t;
78         scanf("%d%d",&n,&m);
79         scanf("%d%d",&s,&t);
80         for(int i=1;i<=m;i++){
81             int u,v,w;
82             scanf("%d%d%d",&u,&v,&w);
83             addedge(u-1,v-1,w*1000+1);
84         }
85         int ans=max_flow(s-1,t-1);
86         printf("%d\n",ans%1000);
87     }
88 
89     return 0;
90 }

Psong

J Brute Force Sorting

链表维护当前剩余的点，队列维护可能被炸掉的点。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int MAXN=1e5+5;
 5 
 6 int n;
 7 int a[MAXN];
 8 int in[MAXN];
 9 int vis[MAXN];
10 int pre[MAXN];
11 int Next[MAXN];
12 
13 void solve(int x){
14     pre[Next[x]]=pre[x];
15     Next[pre[x]]=Next[x];
16     vis[x]=1;
17 }
18 
19 int main(){
20     int t;
21     scanf("%d",&t);
22     while(t--){
23         queue<int> q;
24         scanf("%d",&n);
25         for(int i=1;i<=n;i++){
26             scanf("%d",&a[i]);
27             Next[i]=i+1;
28             pre[i]=i-1;
29             vis[i]=0;
30             in[i]=1;
31             q.push(n+1-i);
32         }
33         while(!q.empty()){
34             int u=q.front();
35             q.pop(); in[u]=0;
36             if(vis[u]) continue;
37             int pos=u;
38             if(pre[pos]>=1&&a[pos]<a[pre[pos]]){
39                 if(Next[pos]<=n&&!in[Next[pos]]){
40                     q.push(Next[pos]);
41                     in[Next[pos]]=1;
42                 }
43                 while(pre[pos]>=1&&a[pos]<a[pre[pos]]){
44                     solve(pos);
45                     pos=pre[pos];
46                 }
47                 if(pos>=1) solve(pos);
48             }
49         }
50         int num=0;
51         for(int i=1;i<=n;i++)
52             if(!vis[i]) num++;
53         printf("%d\n",num);
54         for(int i=1;i<=n;i++)
55             if(!vis[i]) printf("%d ",a[i]);
56         printf("\n");
57     }
58 
59     return 0;
60 }

Psong

K A Cubic number and A Cubic Number

立方差公式为(a-b)(a^2+ab+b^2)，所以a,b需要相邻，预处理出所有的值，然后直接查找即可。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 vector<LL> v;
 5 int main(){
 6     int t;
 7     cin>>t;
 8     for(LL i=1;i<=577350;i++){
 9         LL a=i,b=i+1;
10         v.push_back(a*a+a*b+b*b);
11     }
12     sort(v.begin(),v.end());
13     while(t--){
14         LL x;
15         cin>>x;
16         LL id=lower_bound(v.begin(),v.end(),x)-v.begin();
17         if(v[id]==x) cout<<"YES"<<endl;
18         else cout<<"NO"<<endl;
19     }
20 
21 
22     return 0;
23 }

Psong