【题解】Luogu P1607 [USACO09FEB] 庙会班车 Fair Shuttle 贪心

基本算法2-2/3-1

---

 

先按右端点排序再按左端点排序

记录一个空座位的数组$q[]$

每次有新的牛上车之后，对$q$重新排序

贪心的让牛能上就上，显然优

code

#include <bits/stdc++.h>
using namespace std;
namespace gengyf{
#define ll long long
const int maxn=1e5+10;
const int mod=1e9;
inline int read(){
    int f=1,x=0;char s=getchar();
    while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
    while(s>='0'&&s<='9'){x=x*10+s-'0';s=getchar();}
    return f*x;
}
int q[maxn];
int n,k,c,ans;
struct node{
    int l,r,m;
}a[maxn];
bool cmp(node x,node y){
    return x.r==y.r?x.l>y.l:x.r<y.r;
}
bool cmp2(int x,int y){
    return x>y;
}
int main(){
    k=read();n=read();c=read();
    for(int i=1;i<=k;i++){
        a[i].l=read();a[i].r=read();a[i].m=read();
    }
    sort(a+1,a+1+k,cmp);
    for(int i=1;i<=k;i++){
        sort(q+1,q+1+c,cmp2);
        for(int j=1;j<=c&&a[i].m;j++){
            if(q[j]<=a[i].l){
                ans++;q[j]=a[i].r;a[i].m--;
            }
        }
    }
    printf("%d",ans);
    return 0;
}
}
signed main(){
  gengyf::main();
  return 0;
}