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CCC 2016 S4 - Combining Riceballs

CCC 2016 S4 - Combining Riceballs: a C++ solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Depth-first search
Source
CCCSolutions
Length
130 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For CCC 2016 S4 - Combining Riceballs, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 130 lines of C++ from the credited upstream file ccc16s4.cpp.
  • The implementation visibly relies on cached states.
  • 7 loop blocks detected.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2016 S4 - Combining Riceballs · C++C++
Use this to learn the idea, then write your own version.
/* CCC '16 S4 - Combining Riceballs Dan Shan, Oakville Trafalgar High School Date: 2025-06-13 Recursive Dynamic Programming Observation: order of merging doesn't effect the sum 1. Prefix sum array for O(1) range sum queries. 2. 2D memoization array stores whether each interval [l, r] is valid (can be merged). 3. Recursively try to split into 3 parts: left, right, middle (possibly empty). 4. Optimize Each call from O(N^2) to O(N) through two-pointers technique (Optional but speeds up code significantly) 5. If left and right have equal sums, and all 3 parts are valid, then dp[l][r] is valid. Time Complexity: O(N^3)   C/C++ implementation */   #include <stdio.h> #pragma GCC optimize ("Ofast") #define bs 1<<24 // Templates char buf[bs]; char *ptr = buf; void buff(){   fread(buf,1,bs,stdin); } long long scan(){ // Fast input   long long num=0,neg=1;   while((*ptr<'0'||*ptr>'9')&&*ptr!='-')++ptr; // Skip non-digit characters   while(*ptr=='-')++ptr,neg*=-1;   while(*ptr>='0'&&*ptr<='9') {     num=num*10+(*ptr-'0');     ++ptr;   }   return num*neg; } int dp[401][401],p[401]; int solve(int l, int r){   if(dp[l][r]) return dp[l][r];   if(l>=r) return dp[l][r]=1;   int i=l,j=r;   while(i<=j){     int li=p[i]-p[l-1],ri=p[r]-p[j-1];     if(li!=ri){ // sums don't match       if(li<ri) ++i;       else --j;       continue;     }     if(solve(l,i)>0&&solve(i+1,j-1)>0&&solve(j,r)>0) return dp[l][r]=1;     ++i; --j;   }   return dp[l][r]=-1; } int main(){   buff();   int n=scan(),m=1; p[0]=0;   for(int i=1;i<=n;++i) {     p[i]=scan(); p[i]+=p[i-1];   }   for(int i=1;i<=n;++i){     for(int j=i;j<=n;++j){       int d=p[j]-p[i-1];       if(d>m&&solve(i,j)>0) m=d;     }   }   printf("%d\n",m); }

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