Necklace 单调队列

谁借莪１个温暖的怀抱￠ 2022-01-05 15:29 235阅读 0赞

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

\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*

单调队列

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1 #include<iostream>
     2 #include<string>
     3 #include<cstring>
     4 #include<cstdio>
     5 #include<cmath>
     6 using namespace std;
     7 char str[1500001];
     8 int num[1200001];
     9 int que[1200001];
    10 bool f1[2100001],f2[2100001];
    11 int sum[2100001],n,result;
    12 void  solve(bool t)//单调队列，求最优值
    13 {
    14     int head=1,tail=0;
    15     int it,jt;
    16     for(it=1;it<n;it++)
    17     {
    18         while(head<=tail&&sum[que[tail]]<=sum[it])tail--;
    19         tail++;
    20         que[tail]=it;
    21     }
    22     for(it=n;it<=2*n;it++)
    23     {
    24         while(head<=tail&&sum[que[tail]]<=sum[it])tail--;
    25         tail++;
    26         que[tail]=it;
    27         while(head<=tail&&que[head]<it-n)head++;
    28         if(sum[it]-sum[que[head]]>=0)
    29         {
    30             if(t)
    31               f1[it-n]=true;
    32             else
    33                f2[it-n]=true;
    34         }
    35 
    36     }
    37 }
    38 int main()
    39 {
    40     int cas,T=0,i,j;
    41     scanf("%d",&cas);
    42     while(cas--)
    43     {
    44         T++;
    45         scanf("%s",str+1);
    46         n=strlen(str+1);
    47         memset(f1,false,sizeof(f1));
    48         memset(f2,false,sizeof(f2));
    49         sum[0]=0;
    50         for(i=1;i<=n;i++)
    51         {
    52             if(str[i]=='C')
    53                 num[i]=1;
    54             else
    55                 num[i]=-1;
    56         }
    57         for(i=1;i<=n;i++)
    58          num[i+n]=num[i];
    59         for(i=1;i<=2*n;i++)
    60          sum[i]=sum[i-1]+num[i];
    61         solve(true);
    62         for(i=1;i<=n;i++)
    63          sum[i]=sum[i-1]+num[n+1-i];
    64         for(i=n+1;i<=2*n;i++)
    65           sum[i]=sum[i-1]+num[2*n+1-i];
    66         solve(false);
    67         result=0;
    68         for(i=0;i<n;i++)
    69          if(f1[i]||f2[n-i])
    70           result++;
    71         printf("Case %d: %d\n",T,result);
    72     }
    73     return 0;
    74 }

转载于:https://www.cnblogs.com/sdau--codeants/p/3451706.html

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