Swordsman

ps：比赛的时候想到了做法，k次排序，然后每次消去能消的。。。然而这种做法是错误的，神奇的是测试案例中排在奇数的案例会WA，排在偶数的案例都过了，被注释的代码会T.

#include<bits/stdc++.h>
#define ULL unsigned long long
#define LL long long
#define P pair<int, int>
#define pb push_back
#define mp make_pair
#define pp pop_back
#define lson root << 1
#define INF32 (int)2e9 + 7
#define rson root << 1 | 1
#define INF64 (unsigned long long)1e18
#define sc(x) scanf("%d", &x)
#define pr(x) printf("%d\n", x)
#define mem(arry, in) memset(arry, in, sizeof(arry))
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
using namespace std;

namespace fastIO {
    #define BUF_SIZE 100000
    bool IOerror = 0;
    inline char nc() {
        static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
        if(p1 == pend) {
            p1 = buf;
            pend = buf + fread(buf, 1, BUF_SIZE, stdin);
            if(pend == p1) {
                IOerror = 1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
    }
    inline void read(int &x) {
        char ch;
        while(blank(ch = nc()));
        if(IOerror) return;
        for(x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');
    }
    #undef BUF_SIZE
};
using namespace fastIO;

inline void upd(int &x, int y) { x < y && (x = y); }

const int N = 100005;

P a[15][N];
int T, n, k, v[10], b[N][15], point[15], cnt[N], c[N][15];
bool use[N];

int main()
{
    //freopen("D:\\1.in", "r", stdin);
    //freopen("D:\\1.txt", "w", stdout);
    read(T);
    while(T--) {
        read(n), read(k);
        for (int i = 1; i <= k; ++i) read(v[i]);
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                read(a[j][i].first);
                a[j][i].second = i;
                //c[i][j] = a[j][i].first;
            }
            for (int j = 1; j <= k; ++j) read(b[i][j]);
        }

        mem(cnt, 0);
        mem(point, 0);
        //mem(use, 0);

        int ans = 0;
        for (int i = 1; i <= k; ++i) sort(a[i] + 1, a[i] + 1 + n);
        /*
            //  写的代码真的很搓，也许换种优秀的写法就过了
            sort(a[1] + 1, a[1] + n + 1);
            while(1) {
                int oldans = ans;
            for (int j = 1; j <= n; ++j) if (!use[a[1][j].second]) {
                int id = a[1][j].second;
　　　　　　　　　 if (c[id][1] > v[1]) break;
                int tot = 0;
                for (int q = 1; q <= k; ++q) if (c[id][q] <= v[q]) {
                    tot++;
                }

                if (tot == k) {
                    for (int q = 1; q <= k; ++q) v[q] += b[id][q];
                    use[id] = 1;
                    ans++;
                }
            }
                if (oldans == ans) break;
            }
        
        */
        while(1) {
            int oldans = ans;
            for (int i = 1; i <= k; ++i) {
                while(point[i] < n && a[i][point[i] + 1].first <= v[i]) {
                    int x = a[i][++point[i]].second;
                    cnt[x]++;
                    if (cnt[x] == k) {
                        ++ans;
                        for (int j = 1; j <= k; ++j) v[j] += b[x][j];
                    }
                }
            }
            if (oldans == ans) break;
        }

        pr(ans);
        for (int i = 1; i < k; ++i) printf("%d ", v[i]);
        pr(v[k]);
    }
    return 0;
}