hdu 3474 Necklace 单调队列 DP

Problem Description

You are given a necklace consists of N beads linked as a circle. Each bead is either crystal or jade.
Now, your task is:
1.  Choose an arbitrary position to cut it into a chain.
2.  Choose either direction to collect it.
3.  Collect all the beads in the chosen direction under the constraint that the number of crystal beads in your hand is not less than the jade at any time.
Calculate the number of ways to cut meeting the constraint

 

Input

In the first line there is an integer T, indicates the number of test cases. (T<=50)
Then T lines follow, each line describes a necklace. ‘C’ stands for a crystal bead and ‘J’ stands for a jade bead. The length of necklace is between 2 and 10^6.

 

Output

For each case, print “Case x: d” on a single line in which x is the number of case counted from one and d is the number of ways.

 

Sample Input

  
  

   
   2
CJCJCJ
CCJJCCJJCCJJCCJJ
  
  

 

Sample Output

  
  

   
   Case 1: 6
Case 2: 8
  
  

 

Author

love8909

//

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
/*首先计算从开始到i（0~2*len）点的C与J的数量之差sum[i]根据当
sum[i-len]<=Min(sum[j]|i-len<j<i)时符合题意要求*/
const int maxn=1000010;
char str[maxn*2];
int sum[maxn*2];//从1开始
int cnt;//answer
int len;
bool legal[2][maxn];
int Q[maxn*2];//单调队列
int I[maxn*2];//I[i]表示队列中的Q[i]在数组中的下标
void solve(int index)
{
    int n=strlen(str+1);
    sum[0]=0;
    for(int i=1;i<=n;i++)
    {
        sum[i]=sum[i-1]+(str[i]=='C'?1:-1);
    }
    int head=1,tail=0;
    for(int i=0;i<len;i++) //a[i]入队
    {
        while(head<=tail&&Q[tail]>=sum[i]) tail--;
        tail++;
        Q[tail]=sum[i],I[tail]=i;
    }
    for(int i=len;i<n;i++)
    {
        while(head<=tail&&Q[tail]>=sum[i]) tail--;
        tail++;
        Q[tail]=sum[i],I[tail]=i;
        while(I[head]<=i-len) head++;
        int rmin=Q[head];
        legal[index][i-len]=sum[i-len]<=rmin;
        //if(sum[i-len]<=rmin) cout<<index<<"..."<<i-len<<endl;
    }
    /*
    int head=1,tail=0;
    while(head<=tail&&Q[tail]>=a[i]) tail--;//a[i]入队
    tail++;
    Q[tail]=a[i],I[tail]=i;
    while(I[head]<=i-k) head++;//a[i-k]出队
    _min[i]=Q[head];//保存
    */
}
void _strev(char* str)
{
    for(int i=0,j=strlen(str)-1;i<j;i++,j--)
    {
        char t=str[i];
        str[i]=str[j];
        str[j]=t;
    }
}
int main()
{
    int ci,pl=1;scanf("%d",&ci);
    while(ci--)
    {
        scanf("%s",str+1);
        len=strlen(str+1);
        for(int i=len+1;i<=len+len;i++) str[i]=str[i-len];
        str[len+len+1]='\0';
        memset(legal,0,sizeof(legal));
        solve(0);
        _strev(str+1);
        solve(1);
        cnt=0;
        for(int i=0;i<len;i++)
        {
            if(legal[0][i]||legal[1][(len-i)%len])
            {
                cnt++;
            }
        }
        printf("Case %d: %d\n",pl++,cnt);
    }
    return 0;
}