Codeforces Round #419 (Div. 2) C.Karen and Game 思维

传送门

题意：

给你一个矩阵 问你是不是能化为0矩阵。 每次可以使列或者行减少1。 最后输出操作次数并输出过程。

做法：

取列行最小的一个数。 整个列行都减这个数就好啦。判断一下减了多少次（(s != cc * n + rr * m)）

具体看代码就知道啦。

#include <iostream>#include <stdio.h>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <cmath>#include <map>#include <bitset>#include <set>#include <vector>#include <functional>using namespace std;const int N = 105;int n, m;int A[N][N];int r[N], c[N];int doRow() {  int ans = 0;  for (int i = 0; i < n; ++i)   {    int miv = 1000;    for (int j = 0; j < m; ++j)      miv = std::min(miv, A[i][j]);    r[i]=miv;    ans += r[i];   // cout<<ans<<endl;    for (int j = 0; j < m; ++j)      A[i][j] -= r[i];  }  return ans;}int doCol() {  int ans = 0;  for (int j = 0; j < m; ++j)   {    int miv = 1000;    for (int i = 0; i < n; ++i)      miv = std::min(miv, A[i][j]);    ans += (c[j] = miv);    for (int i = 0; i < n; ++i)      A[i][j] -= c[j];  }  return ans;}int main() {  scanf("%d%d", &n, &m);  int s = 0;  for (int i = 0; i < n; ++i)    for (int j = 0; j < m; ++j)     {      scanf("%d", &A[i][j]);      s += A[i][j];    }  int cc, rr;  if (n > m)   {    cc = doCol();    rr = doRow();  }   else   {    rr = doRow();    cc = doCol();  } // cc=doCol(); // rr=doRow();  if (s != cc * n + rr * m) return puts("-1"), 0;  printf("%d\n", cc + rr);  for (int i = 0; i < n; ++i)    while (r[i]--) printf("row %d\n", i + 1);  for (int j = 0; j < m; ++j)    while (c[j]--) printf("col %d\n", j + 1);  return 0;}