Ivan is a student at Berland State University (BSU). There are n days in Berland week, and each of these days Ivan might have some classes at the university.

There are m working hours during each Berland day, and each lesson at the university lasts exactly one hour. If at some day Ivan's first lesson is during i-th hour, and last lesson is during j-th hour, then he spends j - i + 1 hours in the university during this day. If there are no lessons during some day, then Ivan stays at home and therefore spends 0 hours in the university.

Ivan doesn't like to spend a lot of time in the university, so he has decided to skip some lessons. He cannot skip more than k lessons during the week. After deciding which lessons he should skip and which he should attend, every day Ivan will enter the university right before the start of the first lesson he does not skip, and leave it after the end of the last lesson he decides to attend. If Ivan skips all lessons during some day, he doesn't go to the university that day at all.

Given n, m, k and Ivan's timetable, can you determine the minimum number of hours he has to spend in the university during one week, if he cannot skip more than k lessons?

The first line contains three integers n, m and k (1 ≤ n, m ≤ 500, 0 ≤ k ≤ 500) — the number of days in the Berland week, the number of working hours during each day, and the number of lessons Ivan can skip, respectively.

Then n lines follow, i-th line containing a binary string of m characters. If j-th character in i-th line is 1, then Ivan has a lesson on i-th day during j-th hour (if it is 0, there is no such lesson).

Print the minimum number of hours Ivan has to spend in the university during the week if he skips not more than k lessons.

In the first example Ivan can skip any of two lessons during the first day, so he spends 1 hour during the first day and 4 hours during the second day.

In the second example Ivan can't skip any lessons, so he spends 4 hours every day.

题意 ： 输入数据有 n 行，代表天数，然后每天有 m 节课， 他可以选择性的翘 K 节课，问最终在校时间最短是多少。

思路分析 ： 最开始想的是 3 层 for 的大暴力，每次逃的课一定是某一天的最左或最右的课，但是这样不一定对，可以想一想

后来看了大佬的博客，想明白了，是一个分组背包的问题，我们要做的就是提前预处理出在某一天我们逃 x 节课的所减少的在校时间，当一天的课全部逃后，那么所减少的在校时间就是 m ，然后 3层 for 搞定dp 就ok了

代码示例 ：

int n, m, p;

char s[505];

int pos[505][505], num[505];

int c[505][505]; // 删除 k 个字符后的花费最多可以节约的花费

int dp[505];

int main() {

    //freopen("in.txt", "r", stdin);

    //freopen("out.txt", "w", stdout);

    cin >> n >> m >> p;

    for(int i = 1; i <= n; i++){

        scanf("%s", s+1);

        int pp=1;

        for(int j = 1; j <= m; j++){

            if (s[j] == '1') {num[i]++, pos[i][pp++] = j;}

        }

    }

    for(int i = 1; i <= n; i++){

        for(int j = 0; j <= num[i]-1; j++){

            int res = inf;

            for(int k = 0; k <= num[i]-1; k++){

                if (j-k < 0) break;

                int q = j - k; // q右边开始选多少个1 k表示左边

                res = min(res, pos[i][num[i]-q]-pos[i][k+1]+1);

            }

            c[i][j] = m-res;

        }

        c[i][num[i]] = m;

    }

    //for(int i = 1; i <= n; i++){

        //for(int j = 0; j <= m; j++){

            //cout << c[i][j] << " " ;

        //}

        //cout << endl;

    //}

    for(int i = 1; i <= n; i++){

        for(int j = p; j >= 0; j--){

            for(int k = 0; k <= min(num[i], j); k++){

                dp[j] = max(dp[j], dp[j-k]+c[i][k]);

            }

        }

        //printf("i = %d  dp = %d\n",i, dp[p]);

    }

    cout << n*m-dp[p] << endl;

    return 0;

}