2017 ACM/ICPC Asia Regional Qingdao Online

Posted 2020-10-07 BobHuang

The Dominator of Strings

Time Limit: 3000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 6778    Accepted Submission(s): 713

Problem Description

Here you have a set of strings. A dominator is a string of the set dominating all strings else. The string S is dominated by T if S is a substring of T.

 

 

Input

The input contains several test cases and the first line provides the total number of cases.
For each test case, the first line contains an integer N indicating the size of the set.
Each of the following N lines describes a string of the set in lowercase.
The total length of strings in each case has the limit of 100000.
The limit is 30MB for the input file.

 

 

Output

For each test case, output a dominator if exist, or No if not.

 

 

Sample Input

3

10

you

better

worse

richer

poorer

sickness

health

death

faithfulness

youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness

5

abc

cde

abcde

abcde

bcde

3

aaaaa

aaaab

aaaac

 

 

Sample Output

youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness

abcde

No

从较长串中找子串的

自动机的复杂度太高，挑战上的代码就可以过得

#include<bits/stdc++.h>
  
using namespace std;  
const int MAXN = 100010;  
int n,k;  
int Paiming[MAXN+1],tmp[MAXN+1];  
int flag;  
bool comp_sa(int i, int j)  
{  
    if(Paiming[i] != Paiming[j])   
        return Paiming[i] < Paiming[j];  
    else{
        int ri = i+k <= n? Paiming[i+k] : -1;  
        int rj = j+k <= n? Paiming[j+k] : -1;  
        return ri < rj;     
    }  
}  
  
void calc_sa(string &S, int *sa) 
{  
    n = S.size();  
    for(int i = 0; i <= n; i++)  
    {  
        sa[i] = i;  
        Paiming[i] = i < n ? S[i] : -1;  
    }  
      
    for( k =1; k <= n; k *= 2)  
    {  
        sort(sa,sa+n+1,comp_sa);  
        tmp[sa[0]] = 0;  
        for(int i = 1; i <= n; i++)  
        {  
            tmp[sa[i]] = tmp[sa[i-1]] + (comp_sa(sa[i-1],sa[i]) ? 1: 0);  
        }  
        for(int i = 0; i <= n; i++)  
        {  
            Paiming[i] = tmp[i];  
        }  
    }  
}  
  
int SuffixArrayMatch(string &S, int *sa, string T)  
{  
    
    int lhs = 0, rhs = S.size();  
    while(rhs - lhs > 1)  
    {  
        int mid = (lhs + rhs)>>1;  
        if(S.compare(sa[mid],T.size(),T) < 0) lhs = mid;
        else rhs=mid;   
    }
    return S.compare(sa[rhs],T.size(),T) == 0;  
}  
int main()  
{  
    int t;
    ios::sync_with_stdio(false);
    cin>>t;        
    string s[100010],longs;    
    while(t--){
        int n,l=-1,p=-1,i; 
        cin>>n;
        memset(Paiming,0,sizeof Paiming);
        memset(tmp,0,sizeof tmp);
        for(i=0;i<n;i++){
            cin>>s[i];
            int len=s[i].size();
            if(len>l){
                l=len;
                longs=s[i];
                p=i;
            }
        }
        if(n==1){
            cout<<longs<<endl;
            continue;
        }   
        int *sa = new int[longs.size()+1];  
        calc_sa(longs,sa);
        for(i=0;i<n;i++){
            if(p==i)continue;    
            if(!SuffixArrayMatch(longs,sa,s[i]))
            break;  
        } 
        //delete [] sa;  
        sa = NULL; 
        if(i>=n){
            cout<<longs<<endl;
        }else {
            cout<<"No"<<endl;
        }      
    }   
}  

 

Chinese Zodiac

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2451    Accepted Submission(s): 1645

Problem Description

The Chinese Zodiac, known as Sheng Xiao, is based on a twelve-year cycle, each year in the cycle related to an animal sign. These signs are the rat, ox, tiger, rabbit, dragon, snake, horse, sheep, monkey, rooster, dog and pig.
Victoria is married to a younger man, but no one knows the real age difference between the couple. The good news is that she told us their Chinese Zodiac signs. Their years of birth in luner calendar is not the same. Here we can guess a very rough estimate of the minimum age difference between them.
If, for instance, the signs of Victoria and her husband are ox and rabbit respectively, the estimate should be 2 years. But if the signs of the couple is the same, the answer should be 12 years.

 

Input

The first line of input contains an integer T (1≤T≤1000) indicating the number of test cases.
For each test case a line of two strings describes the signs of Victoria and her husband.

 

Output

For each test case output an integer in a line.

 

Sample Input

3

ox rooster

rooster ox

dragon dragon

 

Sample Output

8

4

12

中国生肖，随手模拟就可以了

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
using namespace std;
int main()
{
    char m[12][10]={
        "rat", "ox", "tiger", "rabbit", "dragon", "snake", "horse", "sheep", "monkey", "rooster", "dog" , "pig"
    };
    int t,s;
    scanf("%d",&t);
    while(t--){
        char p[10],q[10];
        scanf("%s%s",&p,&q);
        if(strcmp(p,q)==0){
            puts("12");
        }else {
            int i,pp,qq;
            for(i=0;i<12;i++){
                if(strcmp(m[i],p)==0){
                    pp=i;
                }
                if(strcmp(m[i],q)==0){
                    qq=i;
                }
            }
            s=(pp-qq);
            if(s>0){
                s=s-12;
            }
            printf("%d\n",-s);
        }
    }
} 

 

Smallest Minimum Cut

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3297    Accepted Submission(s): 602

Problem Description

Consider a network G=(V,E) with source s and sink t. An s-t cut is a partition of nodes set V into two parts such that s and t belong to different parts. The cut set is the subset of E with all edges connecting nodes in different parts. A minimum cut is the one whose cut set has the minimum summation of capacities. The size of a cut is the number of edges in the cut set. Please calculate the smallest size of all minimum cuts.

 

Input

The input contains several test cases and the first line is the total number of cases T (1≤T≤300).
Each case describes a network G, and the first line contains two integers n (2≤n≤200) and m (0≤m≤1000) indicating the sizes of nodes and edges. All nodes in the network are labelled from 1 to n.
The second line contains two different integers s and t (1≤s,t≤n) corresponding to the source and sink.
Each of the next m lines contains three integers u,v and w (1≤w≤255) describing a directed edge from node u to v with capacity w.

 

Output

For each test case, output the smallest size of all minimum cuts in a line.

 

Sample Input

2

4 5

1 4

1 2 3

1 3 1

2 3 1

2 4 1

3 4 2

4 5

1 4

1 2 3

1 3 1

2 3 1

2 4 1

3 4 3

Sample Output

2

3

求点s到点t的最小割，这个网上就有一道一样的题吧

dinic是过不去的，要用sap算法

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MAXN 2333
#define MAXM 2333333
struct Edge
{
    int v,next;
    ll cap;
} edge[MAXM];
int head[MAXN],pre[MAXN],cur[MAXN],level[MAXN],gap[MAXN],NV,NE,n,m,vs,vt;
void ADD(int u,int v,ll cap,ll cc=0)
{
    edge[NE].v=v;
    edge[NE].cap=cap;
    edge[NE].next=head[u];
    head[u]=NE++;

    edge[NE].v=u;
    edge[NE].cap=cc;
    edge[NE].next=head[v];
    head[v]=NE++;
}
ll SAP(int vs,int vt)
{
    memset(pre,-1,sizeof(pre));
    memset(level,0,sizeof(level));
    memset(gap,0,sizeof(gap));
    for(int i=0; i<=NV; i++)cur[i]=head[i];
    int u=pre[vs]=vs;
    ll aug=-1,maxflow=0;
    gap[0]=NV;
    while(level[vs]<NV)
    {
loop:
        for(int &i=cur[u]; i!=-1; i=edge[i].next)
        {
            int v=edge[i].v;
            if(edge[i].cap&&level[u]==level[v]+1)
            {
                aug==-1?aug=edge[i].cap:aug=min(aug,edge[i].cap);
                pre[v]=u;
                u=v;
                if(v==vt)
                {
                    maxflow+=aug;
                    for(u=pre[u]; v!=vs; v=u,u=pre[u])
                    {
                        edge[cur[u]].cap-=aug;
                        edge[cur[u]^1].cap+=aug;
                    }
                    aug=-1;
                }
                goto loop;
            }
        }
        int minlevel=NV;
        for(int i=head[u]; i!=-1; i=edge[i].next)
        {
            int v=edge[i].v;
            if(edge[i].cap&&minlevel>level[v])
            {
                cur[u]=i;
                minlevel=level[v];
            }
        }
        if(--gap[level[u]]==0)break;
        level[u]=minlevel+1;
        gap[level[u]]++;
        u=pre[u];
    }
    return maxflow;
}

int main()
{
    int T,u,v,w;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        scanf("%d%d",&vs,&vt);
        NV=n,NE=0;
        memset(head,-1,sizeof(head));
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            ADD(u,v,(ll)w*MAXM+1);
        }
        ll ans=SAP(vs,vt);
        printf("%d\n",ans%MAXM);
    }
    return 0;
}

 

A Cubic number and A Cubic Number

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4947    Accepted Submission(s): 1346

Problem Description

A cubic number is the result of using a whole number in a multiplication three times. For example, 3×3×3=27 so 27 is a cubic number. The first few cubic numbers are 1,8,27,64 and 125. Given an prime number p. Check that if p is a difference of two cubic numbers.

 

Input

The first of input contains an integer T (1≤T≤100) which is the total number of test cases.
For each test case, a line contains a prime number p (2≤p≤1012).

 

Output

For each test case, output ‘YES‘ if given p is a difference of two cubic numbers, or ‘NO‘ if not.

 

Sample Input

10

2

3

5

7

11

13

17

19

23

29

 

Sample Output

NO

NO

NO

YES

NO

NO

NO

YES

NO

这个我是机房第一个做出来的啊，让你看一个数是不是两个数的立方差

但是这个数是质数，因式分解判断另一段就好的

直接二分答案存不存在就好

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
bool la(LL x)
{
    LL l=1,r=1e6+5;
    while(l<=r)
    {
        LL mi=(l+r)/2;
        LL y=mi-1;
        if(mi*mi+y*y+mi*y==x)
            return 1;
        else if(mi*mi+y*y+mi*y<x)
            l=mi+1;
        else r=mi-1;
    }
    return 0;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        LL n;
        scanf("%lld",&n);
        printf("%s\n",la(n)?"YES":"NO");
    }
    return 0;
}