It's a walking tour day in SIS.Winter, so tt groups of students are visiting Torzhok. Streets of Torzhok are so narrow that students have to go in a row one after another.
Initially, some students are angry. Let's describe a group of students by a string of capital letters "A" and "P":
- "A" corresponds to an angry student
- "P" corresponds to a patient student
Such string describes the row from the last to the first student.
Every minute every angry student throws a snowball at the next student. Formally, if an angry student corresponds to the character with index ii in the string describing a group then they will throw a snowball at the student that corresponds to the character with index i+1i+1 (students are given from the last to the first student). If the target student was not angry yet, they become angry. Even if the first (the rightmost in the string) student is angry, they don't throw a snowball since there is no one in front of them.
Let's look at the first example test. The row initially looks like this: PPAP. Then, after a minute the only single angry student will throw a snowball at the student in front of them, and they also become angry: PPAA. After that, no more students will become angry.
Your task is to help SIS.Winter teachers to determine the last moment a student becomes angry for every group.
The first line contains a single integer tt — the number of groups of students (1≤t≤1001≤t≤100). The following 2t2t lines contain descriptions of groups of students.
The description of the group starts with an integer kiki (1≤ki≤1001≤ki≤100) — the number of students in the group, followed by a string sisi, consisting of kiki letters "A" and "P", which describes the ii-th group of students.
For every group output single integer — the last moment a student becomes angry.
1 4 PPAP
1
3 12 APPAPPPAPPPP 3 AAP 3 PPA
4 1 0
In the first test, after 11 minute the state of students becomes PPAA. After that, no new angry students will appear.
In the second tets, state of students in the first group is:
- after 11 minute — AAPAAPPAAPPP
- after 22 minutes — AAAAAAPAAAPP
- after 33 minutes — AAAAAAAAAAAP
- after 44 minutes all 1212 students are angry
In the second group after 11 minute, all students are angry.
找一下最長的連續P序列即可。
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define LL long long 4 #define inf 0x3f3f3f3f 5 #define pii pair<int,int> 6 #define pb push_back 7 #define all(v) (v.begin(),v.end()) 8 #define mp make_pair 9 10 int main() 11 { 12 int n,t; 13 string str; 14 int f[111]; 15 cin>>t; 16 while(t--){ 17 cin>>n>>str; 18 int tmp=0,ans=0; 19 for(int i=n-1;i>=0;--i){ 20 if(str[i]=='A'){ 21 ans=max(ans,tmp); 22 tmp=0; 23 }else{ 24 tmp++; 25 } 26 } 27 cout<<ans<<endl; 28 } 29 return 0; 30 }
codeforces-1287-B
Bees Alice and Alesya gave beekeeper Polina famous card game "Set" as a Christmas present. The deck consists of cards that vary in four features across three options for each kind of feature: number of shapes, shape, shading, and color. In this game, some combinations of three cards are said to make up a set. For every feature — color, number, shape, and shading — the three cards must display that feature as either all the same, or pairwise different. The picture below shows how sets look.
Polina came up with a new game called "Hyperset". In her game, there are nn cards with kk features, each feature has three possible values: "S", "E", or "T". The original "Set" game can be viewed as "Hyperset" with k=4k=4.
Similarly to the original game, three cards form a set, if all features are the same for all cards or are pairwise different. The goal of the game is to compute the number of ways to choose three cards that form a set.
Unfortunately, winter holidays have come to an end, and it's time for Polina to go to school. Help Polina find the number of sets among the cards lying on the table.
The first line of each test contains two integers nn and kk (1≤n≤15001≤n≤1500, 1≤k≤301≤k≤30) — number of cards and number of features.
Each of the following nn lines contains a card description: a string consisting of kk letters "S", "E", "T". The ii-th character of this string decribes the ii-th feature of that card. All cards are distinct.
Output a single integer — the number of ways to choose three cards that form a set.
3 3 SET ETS TSE
1
3 4 SETE ETSE TSES
0
5 4 SETT TEST EEET ESTE STES
2
In the third example test, these two triples of cards are sets:
- "SETT", "TEST", "EEET"
- "TEST", "ESTE", "STES"
N^2遍歷所有得(i,j)然后可以根據(i,j)字符串構造出來第三個T字符串，然后查找一下是否有這個T存在即可，注意最后答案要/3因為會重復出現。
查找時用了一個map，復雜度是O(N2log(N))這個樣子，但是很慢，幾乎跑滿了。
下來測試了下是查找時候得鍋，利用中括號下標查找很慢，換成M.count(T)飛快了就。
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define LL long long 4 #define inf 0x3f3f3f3f 5 #define pii pair<int,int> 6 #define pb push_back 7 #define all(v) (v.begin(),v.end()) 8 #define mp make_pair 9 10 string e[1505]; 11 int n,K; 12 map<string,int>M; 13 14 int main() 15 { 16 ios::sync_with_stdio(false); 17 int S='S'+'E'+'T'; 18 long long ans=0; 19 cin>>n>>K; 20 for(int i=1;i<=n;++i)cin>>e[i],M[e[i]]++; 21 for(int i=1;i<=n;++i){ 22 for(int j=i+1;j<=n;++j){ 23 if(i==j)continue; 24 string T; 25 for(int o=0;o<K;++o){ 26 if(e[i][o]==e[j][o]) T+=e[i][o]; 27 else{ 28 T+=(char)(S-e[i][o]-e[j][o]); 29 } 30 } 31 ans+=M[T]; 32 } 33 }cout<<ans/3<<endl; 34 return 0; 35 }
codeforces-1287-C
Vadim loves decorating the Christmas tree, so he got a beautiful garland as a present. It consists of nn light bulbs in a single row. Each bulb has a number from 11 to nn (in arbitrary order), such that all the numbers are distinct. While Vadim was solving problems, his home Carp removed some light bulbs from the garland. Now Vadim wants to put them back on.
Vadim wants to put all bulb back on the garland. Vadim defines complexity of a garland to be the number of pairs of adjacent bulbs with numbers with different parity (remainder of the division by 22). For example, the complexity of 1 4 2 3 5 is 22 and the complexity of 1 3 5 7 6 4 2 is 11.
No one likes complexity, so Vadim wants to minimize the number of such pairs. Find the way to put all bulbs back on the garland, such that the complexity is as small as possible.
The first line contains a single integer nn (1≤n≤1001≤n≤100) — the number of light bulbs on the garland.
The second line contains nn integers p1, p2, …, pnp1, p2, …, pn (0≤pi≤n0≤pi≤n) — the number on the ii-th bulb, or 00 if it was removed.
Output a single number — the minimum complexity of the garland.
5 0 5 0 2 3
2
7 1 0 0 5 0 0 2
1
In the first example, one should place light bulbs as 1 5 4 2 3. In that case, the complexity would be equal to 2, because only (5,4)(5,4) and (2,3)(2,3) are the pairs of adjacent bulbs that have different parity.
In the second case, one of the correct answers is 1 7 3 5 6 4 2.
總算遇到了自己擅長得東西了，直接dp，f[i][j][k]表示當前處於下標i處，當前位置的數字奇偶性為j，且剩余未放置的數字中有k個是奇數，由於剩余未放置的數字中偶數得個數可以根據現有狀態推得，所以沒必要多開一維了。
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define LL long long 4 #define inf 0x3f3f3f3f 5 #define pii pair<int,int> 6 #define pb push_back 7 #define all(v) (v.begin(),v.end()) 8 #define mp make_pair 9 10 int n,a[111],f[105][2][105],tot[105],pre[105]; 11 int main() 12 { 13 memset(f,inf,sizeof(f)); 14 cin>>n; 15 int ji=0,ou=0,sum=0; 16 for(int i=1;i<=n;++i) { 17 cin>>a[i]; 18 tot[a[i]]++; 19 if(a[i]==0){ 20 pre[i]=pre[i-1]+1; 21 a[i]=-1; 22 }else{ 23 pre[i]=pre[i-1]; 24 a[i]%=2; 25 } 26 } 27 for(int i=1;i<=n;++i){ 28 if(!tot[i]){ 29 if(i%2==0)ou++; 30 else ji++; 31 } 32 }//cout<<ji<<' '<<ou<<endl; 33 sum=ji+ou; 34 if(a[1]!=-1) f[1][a[1]][ji]=0; 35 else{ 36 if(ji>0) f[1][1][ji-1]=0; 37 if(ou>0) f[1][0][ji]=0; 38 } 39 for(int i=1;i<=n;++i){ 40 for(int j=0;j<=1;++j){ 41 for(int k1=0;k1<=ji;++k1){ 42 int k2=sum-k1-pre[i]; 43 if(f[i][j][k1]!=inf){ 44 if(a[i+1]==0){ 45 if(j==0){ 46 f[i+1][0][k1]=min(f[i+1][0][k1],f[i][j][k1]); 47 }else{ 48 f[i+1][0][k1]=min(f[i+1][0][k1],f[i][j][k1]+1); 49 } 50 }else if(a[i+1]==1){ 51 if(j==1){ 52 f[i+1][1][k1]=min(f[i+1][1][k1],f[i][j][k1]); 53 }else{ 54 f[i+1][1][k1]=min(f[i+1][1][k1],f[i][j][k1]+1); 55 } 56 }else{ 57 //ji 58 if(k1>0){ 59 if(j==1){ 60 f[i+1][1][k1-1]=min(f[i+1][1][k1-1],f[i][j][k1]); 61 }else{ 62 f[i+1][1][k1-1]=min(f[i+1][1][k1-1],f[i][j][k1]+1); 63 } 64 } 65 //ou 66 if(k2>0){ 67 if(j==0){ 68 f[i+1][0][k1]=min(f[i+1][0][k1],f[i][j][k1]); 69 }else{ 70 f[i+1][0][k1]=min(f[i+1][0][k1],f[i][j][k1]+1); 71 } 72 } 73 } 74 } 75 } 76 } 77 } 78 //cout<<f[2][1][1]<<endl; 79 int ans=inf; 80 for(int k1=0;k1<=ji;++k1)ans=min(ans,min(f[n][0][k1],f[n][1][k1])); 81 cout<<ans<<endl; 82 return 0; 83 }
codeforces-1287-D
Evlampiy was gifted a rooted tree. The vertices of the tree are numbered from 11 to nn. Each of its vertices also has an integer aiai written on it. For each vertex ii, Evlampiy calculated cici — the number of vertices jj in the subtree of vertex ii, such that aj<aiaj<ai.
Illustration for the second example, the first integer is
aiai and the integer in parentheses is cici
After the new year, Evlampiy could not remember what his gift was! He remembers the tree and the values of cici, but he completely forgot which integers aiai were written on the vertices.
Help him to restore initial integers!
The first line contains an integer nn (1≤n≤2000)(1≤n≤2000) — the number of vertices in the tree.
The next nn lines contain descriptions of vertices: the ii-th line contains two integers pipi and cici (0≤pi≤n0≤pi≤n; 0≤ci≤n−10≤ci≤n−1), where pipi is the parent of vertex ii or 00 if vertex ii is root, and cici is the number of vertices jj in the subtree of vertex ii, such that aj<aiaj<ai.
It is guaranteed that the values of pipi describe a rooted tree with nn vertices.
If a solution exists, in the first line print "YES", and in the second line output nn integers aiai (1≤ai≤109)(1≤ai≤109). If there are several solutions, output any of them. One can prove that if there is a solution, then there is also a solution in which all aiai are between 11 and 109109.
If there are no solutions, print "NO".
3 2 0 0 2 2 0
YES 1 2 1
5 0 1 1 3 2 1 3 0 2 0
YES 2 3 2 1 2
借鑒了https://blog.csdn.net/qq_45458915/article/details/103856822得思路。
假設當前節點得所有兒子已經處理好為了一個數列，數列的下標代表這個元素的rank，根據當前的c[u]就可以確定應該把u插入到這個序列的哪個位置。
所以只要有解的話用1-n這n個數恰好能構成一個解，題目把范圍說那么大純屬唬人>_<
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define LL long long 4 #define inf 0x3f3f3f3f 5 #define pii pair<int,int> 6 #define pb push_back 7 #define all(v) (v.begin(),v.end()) 8 #define mp make_pair 9 10 const int maxn=2010; 11 int n,c[maxn],r[maxn],p,root,ok=1; 12 vector<int>son[maxn]; 13 vector<int> dfs(int u){ 14 vector<int>res; 15 for(auto v:son[u]){ 16 vector<int>tp=dfs(v); 17 res.insert(res.end(),tp.begin(),tp.end()); 18 } 19 20 if(c[u]>res.size()){cout<<"NO"<<endl;exit(0);} 21 res.insert(res.begin()+c[u],u); 22 return res; 23 } 24 int main() 25 { 26 cin>>n; 27 for(int i=1;i<=n;++i){ 28 cin>>p>>c[i]; 29 if(p==0)root=i; 30 son[p].pb(i); 31 } 32 vector<int>res = dfs(root); 33 cout<<"YES"<<endl; 34 for(int i=0;i<n;++i)r[res[i]]=i+1; 35 for(int i=1;i<=n;++i)cout<<r[i]<<' ';cout<<endl; 36 return 0; 37 }