Problem:
Score : 500 points

Problem Statement

We have a chocolate bar partitioned into H horizontal rows and W vertical columns of squares.

The square (i, j) at the i-th row from the top and the j-th column from the left is dark if S_{i,j} is 0, and white if S_{i,j} is 1.

We will cut the bar some number of times to divide it into some number of blocks. In each cut, we cut the whole bar by a line running along some boundaries of squares from end to end of the bar.

How many times do we need to cut the bar so that every block after the cuts has K or less white squares?

Constraints

1 \leq H \leq 10

1 \leq W \leq 1000

1 \leq K \leq H \times W

S_{i,j} is 0 or 1.

Input

Input is given from Standard Input in the following format:

H W K
S_{1,1}S_{1,2}...S_{1,W}
:
S_{H,1}S_{H,2}...S_{H,W}

Output

Print the number of minimum times the bar needs to be cut so that every block after the cuts has K or less white squares.

Sample Input 1

3 5 4
11100
10001
00111

Sample Output 1

2

For example, cutting between the 1-st and 2-nd rows and between the 3-rd and 4-th columns - as shown in the figure to the left - works.

Note that we cannot cut the bar in the ways shown in the two figures to the right.

Sample Input 2

3 5 8
11100
10001
00111

Sample Output 2

0

No cut is needed.

Sample Input 3

4 10 4
1110010010
1000101110
0011101001
1101000111

Sample Output 3

3

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <cmath>

using namespace std;
vector<vector<bool> > graph;

const long long INF = 1LL << 61;


int main(){
    int H, W, K;
    cin >> H >> W >> K;

    graph.assign(H, vector<bool>());


    for(int h=0; h<H; ++h){
        string s;
        cin >> s;
        for(int w=0; w<W; ++w){
            int flag = s[w] - '0';
            graph[h].push_back(flag);
        }
    }

    //答え
    long long ans = INF;

    //横の切り方はbit全探索2^H通り
    for(int bit=0; bit< (1 << (H-1)); ++bit){
        int group = 0;
        vector<int> id(H);
        //どのグループに属するかを決める
        for(int h=0; h<H; ++h){
            id[h] = group;
            if(bit >> h & 1){
                group++;
            }
        }
        group++;

        //各グループに今何個入っているか
        vector<vector<int>> group_nums(group, vector<int>(W));
        for(int h=0; h<H; ++h){
            for(int w=0; w<W; ++w){
                int group = id[h];
                if(graph[h][w] == true){
                    group_nums[group][w]++;
                }
            }
        }


        //縦に何回切るかを貪欲に計算
        bool skip = false;
        vector<vector<int>> group_cumsum(group, vector<int>(W));
        //縦の累積を初期化
        for(int i=0; i<group; ++i){
            group_cumsum[i][0] = group_nums[i][0];
        }

        for(int i=0; i<group; ++i){
            for(int w=0; w<W; ++w){
                if(graph[i][w] > K){
                    skip = true;
                }
            }
        }


        if(!skip){
            long long tate = 0;
            int w=1;
            //一個ずつ切るか見ていく
            while(w <= W-1){
                bool is_cut = false;
                for(int i=0; i<group; ++i){
                    long long next_num = group_cumsum[i][w-1] + group_nums[i][w];
                    if(next_num > K){
                        is_cut = true;
                    }
                }

                if(is_cut == true){
                    //カットする
                    tate++;
                    //そうすると次の累積和はgraph[g][w]に初期化される.
                    for(int i=0; i<group; ++i){
                        group_cumsum[i][w] = group_nums[i][w];
                        
                    }                    
                }else{
                    //カットしない場合は累積和を増やす
                    for(int i=0; i<group; ++i){
                        group_cumsum[i][w] = group_cumsum[i][w-1] + group_nums[i][w];
                        
                    }
                }
                w++;
            }

            long long cut_num = tate + (group-1);
            ans = min(ans, cut_num);
        }
    }

    cout << ans << endl;
}