codeforces#FF(div2) D DZY Loves Modification_html/css_WEB-ITnose

程序员文章站 2022-05-12 11:35:51

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首先要知道选择行列操作时顺序是无关的

用两个数组row[i],col[j]分别表示仅选择i行能得到的最大值和仅选择j列能得到的最大值

这个用优先队列维护，没选择一行(列)后将这行(列)的和减去相应的np (mp)重新加入队列

枚举选择行的次数为i,那么选择列的次数为k - i次，ans = row[i] + col[k - i] - (k - i) * i * p;

既然顺序无关，可以看做先选择完i次行，那么每次选择一列时都要减去i * p，选择k - i次列，即减去(k - i) * i * p

//#pragma comment(linker, "/STACK:102400000,102400000")//HEAD#include #include #include #include #include #include #include #include #include #include #include #include using namespace std;//LOOP#define FE(i, a, b) for(int i = (a); i = (a); --i)#define REP(i, N) for(int i = 0; i = (a); --i)#define CPY(a, b) memcpy(a, b, sizeof(a))#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)#define EQ(a, b) (fabs((a) - (b))  VI;typedef unsigned long long ULL;typedef long long LL;const int INF = 0x3f3f3f3f;const int maxn = 1010;const double eps = 1e-10;const LL MOD = 1e9 + 7;int ipt[maxn][maxn];LL row[maxn * maxn], col[maxn * maxn];LL rtol[maxn], ctol[maxn];int main(){    int n, m, k, p;    while (~RIV(n, m, k, p))    {        priority_queue r, c;        int radd = 0, cadd = 0;        CLR(rtol, 0), CLR(ctol, 0);        FE(i, 1, n)            FE(j, 1, m)            {                RI(ipt[i][j]);                rtol[i] += ipt[i][j];                ctol[j] += ipt[i][j];            }        FE(i, 1, n)            r.push(rtol[i]);        FE(j, 1, m)            c.push(ctol[j]);        row[0] = 0, col[0] = 0;        FE(i, 1, k)        {            LL x = r.top(), y = c.top();             r.pop(), c.pop();            r.push(x - m * p);            c.push(y - n * p);            row[i] = row[i - 1] + x;            col[i] = col[i - 1] + y;        }//        FE(i, 0, k)//            cout