HDU6387 AraBellaC

题意

AraBellaC

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 277    Accepted Submission(s): 100

Problem Description

Princess Arabella loves the letter `C`, which earned her the epithet `ArraBellaC`. She would like to endeavor to construct a periodic string with length n(n≤20000) to spare her tedious time. In every period of the string, the substring begins with dozens of `A`s and ends with dozens of `C`s, and dozens of `B`s are inserted between them, and the number of `A`, `B`, `C` is a, b, c.(For example, if a=1, b=1, c=2 then one of the period of the string is `ABCC`, and the periodic string is `ABCCABCC……`).Note that in the last period of the string, the substring can be incomplete and the last few letters of the period can be cut down.

And now, Arabella has constructed a periodic string, she want to give you a question: if she tells you the character is Ci(Ci∈{A,B,C}) in the Xi(Xi≤10000) , then could you tell her the value of a,b,c? If there are multiple answers, please find the lexicographically smallest answer. If you can’t find a valid answer, please print `NO`

 

Input

The first line gives an integer T(1≤T≤40) , which indicates the number of cases in the input.
The first line of every case contains only one integer m(1≤m≤5000), which means the number of characters AraBella told you.
A number Xi(1≤Xi≤10000) and a letter Ci are given in the following m lines, which means the position Xi in the string with letter Ci.

 

Output

Print a,b,c in order. If there are multiple answers, please find the lexicographically smallest answer. If you can’t find a valid answer, please print `NO`.(It is gratuated that 0<a,b,c, and please forget my poor Yinglish)

 

Sample Input

2
3
1 A
2 B
3 C
4
1 A
2 C
3 B
4 C

 

Sample Output

1 1 1
NO

 

Source

2018 Multi-University Training Contest 7

 

Recommend

chendu

 

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分析

参照616156的题解。

题解又在卖萌了。。要用RMQ？。。。估计他自己没想清楚吧。。。这题哪用得着RMQ啊。。。

首先，可以暴力枚举循环节的长度：
然后，对每种字母进行判断，求出每个循环节中B,C最靠前的位置和A,B最靠后的位置。
如果A最靠后的位置在B最靠前的后面，或者B最靠后的位置在C最靠前的后面。就说明当前这个循环节长度是矛盾的。

如果不矛盾，则将A最靠后的位置及以前，都设为A，B最靠后的位置及以前（除去A的部分）都设为B，C则填充这个循环节剩余的位置。就能保证是当前长度循环节中，答案字典序最小一种方案。

问题就在于如何处理:“每个循环节中B,C最靠前的位置和A,B最靠后的位置”
可以利用调和级数：暴力枚举每个循环节，O(1)求出当前循环节最靠前的B、C，最靠后的A、B。总的时间复杂度就是O(N ln N)。

题解的方法说要用RMQ求这个问题。。。
其实就用前缀和后缀的思想不就行了？

对于每个点，它前面最近的一个A、B字母的位置，以及它后面最靠近的一个B、C字母的位置。这两个东西可以O(N)递推求出来

询问某个循环节时，求出其左端点及以后，最靠前的B、C，以及其右端点及以前最靠后的A、B即可（就是上面维护的那两个东西）。

代码

#include<bits/stdc++.h>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;
    rg char ch=getchar();
    while(!isdigit(ch)){
        if(ch=='-') w=-1;
        ch=getchar();
    }
    while(isdigit(ch))
        data=data*10+ch-'0',ch=getchar();
    return data*w;
}
template<class T>il T read(rg T&x){
    return x=read<T>();
}
typedef long long ll;
using namespace std;
co int N=1e4;
int a[N+10],pre[N+10][4],las[N+10][4],ans[4],ans1[4];
char s[20];
int get_pre(int len,int x){
    int res=len;
    for(int i=1;i<=N;i+=len){
        int t=las[i][x];
        if(t>=i+len) continue;
        res=min(res,t-i+1);
    }
    return res;
}
int get_las(int len,int x){
    int res=0;
    for(int i=1;i<=N;i+=len){
        int top=min(i+len-1,N);
        int t=pre[top][x];
        if(t<i) continue;
        res=max(res,t-i+1);
    }
    return res;
}
int main(){
//  freopen(".in","r",stdin),freopen(".out","w",stdout);
    int kase=read<int>();
    while(kase--){
        int m=read<int>();
        memset(ans,-1,sizeof ans);
        memset(a,0,sizeof a);
        memset(pre,0,sizeof pre);
        memset(las,0x3f,sizeof las);
        for(int i=1,pos;i<=m;++i){
            read(pos);scanf("%s",s);
            a[pos]=s[0]-'A'+1;
        }
        for(int i=1;i<=N;++i)
            for(int j=1;j<=3;++j){
                if(a[i]==j) pre[i][j]=i;
                pre[i][j]=max(pre[i][j],pre[i-1][j]);
            }
        for(int i=N;i>=1;--i)
            for(int j=1;j<=3;++j){
                if(a[i]==j) las[i][j]=i;
                las[i][j]=min(las[i][j],las[i+1][j]);
            }
        for(int len=3;len<=N;++len){
            int lasx=0;
            bool flag=0;
            for(int j=1;j<=3;++j){
                if(j!=1){
                    int prex=get_pre(len,j);
                    if(prex<=lasx) {flag=1;break;}
                }
                if(j!=3){
                    lasx=max(lasx+1,get_las(len,j));
                    ans1[j]=lasx;
                }
                else ans1[j]=len;
            }
            if(flag==0){
                for(int j=1;j<=3;++j){
                    if(ans1[j]<ans[j]||ans[j]==-1){
                        ans[1]=ans1[1],ans[2]=ans1[2],ans[3]=ans1[3];
                        break;
                    }
                    if(ans1[j]>ans[j]) break;
                }
            }
        }
        if(ans[1]!=-1) printf("%d %d %d\n",ans[1],ans[2]-ans[1],ans[3]-ans[2]);
        else puts("NO");
    }
    return 0;
}