正题
题目链接:https://www.luogu.com.cn/problem/CF496E
题目大意
nnn个[li,ri][l_i,r_i][li,ri],mmm个[ai,bi,ki][a_i,b_i,k_i][ai,bi,ki]表示可以覆盖掉kik_iki个ai≤li≤ri≤bia_i\leq l_i\leq r_i\leq b_iai≤li≤ri≤bi的区间,求能否覆盖所有区间。
解题思路
右端点排序后,然后每个优先选取左端点最小的区间即可,用平衡树维护。
时间复杂度O(nlogn)O(n\log n)O(nlogn)
codecodecode
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
using namespace std;
const int N=1e5+10;
struct node{int id,l,r,k;
}a[N],b[N];
multiset<node> s;
int n,m,v[N],ans;
bool operator<(node x,node y){return x.l<y.l;}
bool cmp(node x,node y){return x.r<y.r;}
int main()
{scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d%d",&b[i].l,&b[i].r),b[i].id=i;scanf("%d",&m);for(int i=1;i<=m;i++)scanf("%d%d%d",&a[i].l,&a[i].r,&a[i].k),a[i].id=i;sort(b+1,b+1+n,cmp);sort(a+1,a+1+m,cmp);int z=1;for(int i=1;i<=m;i++){while(z<=n&&b[z].r<=a[i].r)s.insert(b[z++]);set<node>::iterator it=s.lower_bound(a[i]);while(a[i].k&&it!=s.end()){node k=*it;ans++;a[i].k--;v[k.id]=a[i].id;s.erase(it);it=s.lower_bound(a[i]);}} if(ans!=n)printf("NO");else{printf("YES\n");for(int i=1;i<=n;i++)printf("%d ",v[i]); }
}