簡單的樹狀數組,複雜度是O(n+m)logn
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define maxn 100005#define maxm 50005int cal[maxn],ans[maxn],sum[maxm],mid[maxn];int n,m;struct TNode{int l,r,k,id;}nod[maxm];bool cmp(TNode a,TNode b){return a.l<b.l||(a.l==b.l&&a.r<b.r);}int lowbit(int x) {return x&(-x);}void update(int x,int val){for(;x<=n;x+=lowbit(x)) cal[x]+=val;}int getsum(int x){int s=0;for(;x>0;x-=lowbit(x)) s+=cal[x];return s;}int find(int k){int l=1,r=n,ms,mt;while(l<=r){ms=(l+r)>>1;mt=getsum(ms);if(mt>=k) r=ms-1;else l=ms+1;}return l;}void solve(){int l=0,r=0,i,j,k,p,q;memset(cal,0,sizeof(cal));l=nod[0].l;r=nod[0].r;for(i=l;i<=r;i++){k=lower_bound(ans+1,ans+1+n,mid[i])-ans;update(k,1);}sum[nod[0].id]=ans[find(nod[0].k)];for(i=1;i<m;i++){p=nod[i].l-1<r?nod[i].l-1:r;q=nod[i].l-1>r?nod[i].l-1:r;if(nod[i].l>l)for(j=l,l=nod[i].l;j<=p;j++){k=lower_bound(ans+1,ans+1+n,mid[j])-ans;update(k,-1);}if(nod[i].r>r)for(j=q+1,r=nod[i].r;j<=nod[i].r;j++){k=lower_bound(ans+1,ans+1+n,mid[j])-ans;update(k,1);}sum[nod[i].id]=ans[find(nod[i].k)];}for(i=0;i<m;i++) printf("%d\n",sum[i]);}int main(){int i,j,k,r,l;scanf("%d%d",&n,&m);for(i=1;i<=n;i++) {scanf("%d",&ans[i]);mid[i]=ans[i];}sort(ans+1,ans+n+1);n=unique(ans+1,ans+n+1)-(ans+1);for(i=0;i<m;i++){scanf("%d%d%d",&nod[i].l,&nod[i].r,&nod[i].k);nod[i].id=i;}sort(nod,nod+m,cmp);solve();return 0;}