Real problems.
Your next challenge.

Problem summary

Each segment has a length and one of three colors: b, g, r. Count choices of three distinct input segments that form a nondegenerate triangle and use all three colors. Equal-length segments remain separate choices.

Constraints

1 ≤ N ≤ 6000; lengths are positive integers at most 100000.

Worked example

Original statement sample.

5
r 10
g 10
b 12
r 5
g 6

3

Solution analysis

1. Sort every segment by length, keeping its color. For each right endpoint c, count pairs a<b<c whose lengths sum to more than length[c]. Since c is a largest side, this strict inequality is exactly the triangle condition.
2. Use a left pointer a=0 and a middle pointer b=c−1. If length[a]+length[b] is too small, advance a. Otherwise every index from a through b−1 also makes a large-enough sum with b. Only the color different from both color[b] and color[c] is acceptable, and if those two colors agree the contribution is zero. Prefix color counts give the number of acceptable indices in [a,b) in O(1), after which decrement b.
3. In the sample, the blue segment of length 12 can join red/green pairs (10,10), (10,6) and (5,10). Pair (5,6) fails because 5+6≤12, leaving three triangles. Index-based counting preserves multiplicities for equal lengths.

Why it works

At a fixed largest index c, the two-pointer argument partitions all pairs: a failed smallest value cannot work with any smaller b, while a successful b works with every remaining index ≥a. Prefix counts select exactly pairs with three different colors. Each triple has a unique largest sorted index, so it is counted once.

Watch out. Use a strict sum inequality to exclude degenerate triangles. Do not discard duplicate segments. When the two fixed colors match, no third color can make all three distinct.

An independently written complete program with standard headers.

#include<algorithm>
#include<array>
#include<iostream>
#include<vector>
using namespace std;
struct Segment{
    int length;
    int color;
};
int main(){
    int n;
    cin>>n;
    vector<Segment>segments(n);
    for(Segment&segment:segments){
        char color;
        cin>>color>>segment.length;
        segment.color=color=='b'?0:color=='g'?1:2;
    }
    sort(segments.begin(),segments.end(),[](const Segment&a,const Segment&b){
        return a.length<b.length;
    });
    vector<array<int,3>>prefix(n+1);
    for(int i=0;i<n;i++){
        prefix[i+1]=prefix[i];
        prefix[i+1][segments[i].color]++;
    }
    long long answer=0;
    for(int largest=2;largest<n;largest++){
        int left=0;
        int right=largest-1;
        while(left<right){
            if(segments[left].length+segments[right].length<=segments[largest].length){
                left++;
            }
            else{
                if(segments[right].color!=segments[largest].color){
                    int missing=3-segments[right].color-segments[largest].color;
                    // Count the required third color among [left, right).
                    answer+=prefix[right][missing]-prefix[left][missing];
                }
                right--;
            }
        }
    }
    cout<<answer<<'\n';
}