关于贪心题型的总结（没写完）

排座椅：
用到了排序的贪心，由于数量不知道多少，所以先找排完序最优的，可得最优解（一定关注题面上是否有最字）
#include <bits/stdc++.h>
using namespace std;
inline int read(){
    int num=0,f=1; char c=getchar();
    while(!isdigit(c)){if(c=='-') f=-1; c=getchar();}
    while(isdigit(c)){num=(num<<1)+(num<<3)+(c^48); c=getchar();}
    return num*f;
}
const int maxn=2010;
//cautious:to avoid making mistakes please be careful to the range of the data
int m,n,k,l,d,aj[maxn],bj[maxn],a[maxn],b[maxn];
bool jud1[maxn],jud2[maxn];
inline int in(){
    m=read(); n=read(); k=read(); l=read(); d=read();
    for(int i=1;i<=d;i++){
        int x1,y1,x2,y2;
        y1=read(); x1=read(); y2=read(); x2=read();
        if(x1==x2) aj[min(y1,y2)]++;
        if(y1==y2) bj[min(x1,x2)]++;
    }
}
inline int test2(){
    for(int i=1;i<=2010;i++)
       for(int j=1;j<=bj[i];j++)
          if(bj[i])
            cout<<i<<" "<<bj[i]<<endl;
}
//cautious(important):do not try to sort the bucket it will make it valueless 
inline int work(){
    
    
    int maxx,cnt=0,ans=0,num;
    for(int i=1;i<=k;i++){
        maxx=0;
        for(int j=1;j<=2010;j++){
            if(aj[j]>maxx and jud1[j]==0){
//cautious:after the change the jud[j] is changed to the jud[num] without observing that leads to a big mistake
                maxx=aj[j]; num=j;
            }
        }
        jud1[num]=1;
        cnt++;
        a[cnt]=num;
    }
    sort(a+1,a+k+1);
    
    
    for(int i=1;i<=l;i++){
        maxx=0;
        for(int j=1;j<=2010;j++){
            if(bj[j]>maxx and jud2[j]==0){
                //cout<<"maxx:"<<maxx<<" "<<"j:"<<j<<" "<<"bj[j]:"<<bj[j]<<endl;
                maxx=bj[j]; num=j;
            }
        }
        jud2[num]=1;
        ans++;
        b[ans]=num;
    }
    //for(int i=1;i<=l;i++) cout<<b[i]<<" ";
    sort(b+1,b+l+1);
    
    
}
inline int test1(){
    for(int i=1;i<=k;i++) cout<<a[i]<<" ";
    cout<<endl;
    for(int i=1;i<=l;i++) cout<<b[i]<<" ";
}
inline int out(){
    for(int i=1;i<=k;i++){
        if(i!=k) cout<<a[i]<<" ";
        else cout<<a[i]<<endl;
    }
    for(int i=1;i<=l;i++){
        if(i!=l) cout<<b[i]<<" ";
        else cout<<b[i]; 
    }
}
//cautious:there are no any blanks when i==k
int main(){
    freopen("seats.in","r",stdin);
    freopen("seats.out","w",stdout);
    in();
    work();
    out();
    //test1();
    //test2();
}


均分制牌：
本题用到了等效的方法：对于样的有限制的左右两侧传递模型，可以考虑单一方向的传递并考虑负数（很重要的一种思路）

#include <bits/stdc++.h>
using namespace std;
const int maxn=10010;
int n,sum,a[maxn],ans;
inline int read(){
int num=0,f=1; char c=getchar();
while(!isdigit(c)){if(c=='-') f=-1; c=getchar();}
while(isdigit(c)){num=(num<<1)+(num<<3)+(c^48); c=getchar();}
return num*f;
}
inline int in(){
n=read();
for(int i=1;i<=n;i++){
a[i]=read(); sum+=a[i];
}
sum/=n;
}
inline int work(){
for(int i=1;i<=n;i++){
int cnt=a[i]-sum;
if(cnt==0) continue;
else{
a[i+1]=a[i+1]+cnt;
ans++;
}
}
}
inline int out(){
cout<<ans;
}
int main(){
in();
work();
out();
}