冥土追魂（暴力枚举）

刚开始以为是博弈，原来是暴力。

可以证明：最优情况下一定是最多只能有一行只选一部分，剩下n-1行要不整行全选，要不不选

也就是说对于当前k，暴力枚举哪一行选择前(k%m)个，然后剩下n-1行中选择所有和最小的(k/m)行就可以了

复杂度O(n*m*logm+n²)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<set>
#define mem(a,x) memset(a,x,sizeof(a))
#define s1(x) scanf("%d",&x)
#define s2(x,y) scanf("%d%d",&x,&y)
#define s3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s4(x,y,z,k) scanf("%d%d%d%d",&x,&y,&z,&k)
#define ls 2*rt
#define rs 2*rt+1
#define lson ls,L,mid
#define rson rs,mid+1,R
#define ll long long
using namespace std;
typedef pair<int,int> pii;
//inline ll ask(int x){ll res=0;while(x)res+=c[x],x-=x&(-x);return res;}
//inline void add(int x,int d){while(x<=n)c[x]+=d,x+=x&(-x);}
//int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b);}
const ll inf = 0x3f3f3f3f;
const int mx = 1005;
struct no{
  ll a[mx],sum[mx];
}p[mx];
int n,m,k;
bool cmp1(int a, int b){
  return a>b;
}
bool cmp2(no &a, no &b){
  return a.sum[m]<b.sum[m];
}
ll calc(int num, int x){
   ll ans = 0,co = 0;
   for(int i = 1; i <= n&& co < num; i++){
    if(i!=x){
      co++;
      ans += p[i].sum[m];  
    }
   }
   return ans;
}
int main(){
  //  freopen("F:\\in.txt","r",stdin);
  //int T=10; scanf("%d",&T);
  s3(n,m,k);
  for(int i = 1; i <= n ; i++){
    for(int j = 1; j <= m; j++){
      scanf("%lld",&p[i].a[j]);
      //p[i].sum[j] = p[i].sum[j-1]+p[i].a[j];
    }
  }
  for(int i = 1; i <= n ; i++){
    p[i].sum[0] = 0;
    sort(p[i].a+1,p[i].a+1+m,cmp1);
    for(int j = 1; j <= m; j++){
      p[i].sum[j] = p[i].sum[j-1]+p[i].a[j];
    }
  }
  sort(p+1,p+1+n,cmp2);
  int r1 = k/m,r2 = k%m;
  ll ans = 1e17;
  for(int i = 1; i <= n; i++){
    if(r2 == 0){
      ans = min(ans,calc(r1,-1));
    }
    else{
      ans = min(ans,p[i].sum[r2]+calc(r1,i));  
    }
    
  }
  cout<<ans<<endl;
  
  return 0;
}

​​om/acm/contest/212/B​​