1. 815A Karen and Game
大意: 给定$nm$矩阵, 每次选择一行或一列全部减$1$, 求最少次数使得矩阵全$0$
贪心, $n>m$时每次取一列, 否则取一行
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl '\n' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+50; int n,m,a[111][111]; int op[N], p[N], cnt; void add(int tp, int x) { ++cnt; op[cnt] = tp, p[cnt] = x; if (tp==1) REP(i,1,n) --a[i][x]; else REP(i,1,m) --a[x][i]; } int main() { scanf("%d%d",&n,&m); int sum = 0; REP(i,1,n) REP(j,1,m) scanf("%d",a[i]+j),sum+=a[i][j]; while (sum) { if (n>m) { int ok = 0; REP(i,1,m) { int s = 1e9; REP(j,1,n) s=min(s,a[j][i]); if (s) add(1,i),sum-=n,ok=1; } if (!ok) { REP(i,1,n) { int s = 1e9; REP(j,1,m) s=min(s,a[i][j]); if (s) add(2,i),sum-=m,ok=1; } if (!ok) return puts("-1"),0; } } else { int ok = 0; REP(i,1,n) { int s = 1e9; REP(j,1,m) s=min(s,a[i][j]); if (s) add(2,i),ok=1,sum-=m; } if (!ok) { REP(i,1,m) { int s = 1e9; REP(j,1,n) s=min(s,a[j][i]); if (s) add(1,i),ok=1,sum-=n; } if (!ok) return puts("-1"),0; } } } printf("%d\n",cnt); REP(i,1,cnt) printf("%s %d\n",op[i]==1?"col":"row",p[i]); }