題目鏈接: https://www.lydsy.com/JudgeOnline/problem.php?id=2716

怎么KD樹(shù)跑得都那么快啊。。我寫的CDQ分治被暴虐

做四遍CDQ分治，每次求一個(gè)左下角\(x_i+y_i\)的最大值

第一種寫法是一開(kāi)始按時(shí)間排序，然后CDQ分治的時(shí)候改成按\(x\)坐標(biāo)排序，同時(shí)用樹(shù)狀數(shù)組統(tǒng)計(jì)每個(gè)\(y\)坐標(biāo)的最大值

第二種寫法是一開(kāi)始按\(x\)坐標(biāo)排序，然后CDQ分支的時(shí)候改成按時(shí)間排序

CDQ分治好神奇(琦)。。。

一定要注意樹(shù)狀數(shù)組如果沒(méi)有元素不能返回0! 我這么寫然后test1 WA on line 40W+...

代碼

#include<cstdio> #include<cstdlib> #include<cstring> #include<cassert> #include<algorithm> using namespace std;const int N = 1e6; const int C = 1e6+2; struct Query {int opt,x,y,id,ans;bool operator <(const Query &arg) const {return id<arg.id;} } qr[N+3],qr2[N+3]; int tr[C+3]; int n,q,mx;void modify(int lrb,int val) {while(lrb<=mx){tr[lrb] = max(tr[lrb],val);lrb += (lrb&(-lrb));} }int querymax(int rb) {int ret = -C*2;while(rb>0){ret = max(ret,tr[rb]);rb -= (rb&(-rb));}return ret; }void clear(int lrb) {while(lrb<=mx && tr[lrb]!=-C*2){tr[lrb] = -C*2;lrb += (lrb&(-lrb));} }void cdqdc(int lb,int rb) {if(lb==rb) return;int mid = (lb+rb)>>1;int j = lb,k = mid+1;for(int i=lb; i<=rb; i++){if(qr[i].id<=mid) {qr2[j] = qr[i]; j++;}else {qr2[k] = qr[i]; k++;}}for(int i=lb; i<=rb; i++) qr[i] = qr2[i];cdqdc(lb,mid);cdqdc(mid+1,rb);j = lb; k = mid+1;while(k<=rb){while(j<=mid && qr[j].opt==2) j++;while(k<=rb && qr[k].opt==1) k++;if(k>rb) break;while(j<=mid && qr[j].x<=qr[k].x){if(qr[j].opt==1) {modify(qr[j].y,qr[j].x+qr[j].y);}j++;}int tmp = querymax(qr[k].y);qr[k].ans = min(qr[k].ans,qr[k].x+qr[k].y-tmp);k++;}for(int i=lb; i<=mid; i++) {if(qr[i].opt==1) clear(qr[i].y);}j = lb; k = mid+1;for(int i=lb; i<=rb; i++){if(j>mid || (k<=rb && qr[k].x<qr[j].x)) {qr2[i] = qr[k]; k++;}else {qr2[i] = qr[j]; j++;}}for(int i=lb; i<=rb; i++) qr[i] = qr2[i]; }int main() {scanf("%d%d",&n,&q);for(int i=1; i<=n; i++){scanf("%d%d",&qr[i].x,&qr[i].y); qr[i].opt = 1; qr[i].id = i; qr[i].x++; qr[i].y++; qr[i].ans = C*7;mx = max(mx,max(qr[i].x,qr[i].y));}for(int i=n+1; i<=n+q; i++){scanf("%d%d%d",&qr[i].opt,&qr[i].x,&qr[i].y,&qr[i].opt); qr[i].id = i; qr[i].x++; qr[i].y++; qr[i].ans = C*7;mx = max(mx,max(qr[i].x,qr[i].y));}q+=n;for(int i=0; i<=mx; i++) tr[i] = -C*2;cdqdc(1,q);sort(qr+1,qr+q+1);for(int i=1; i<=q; i++) {qr[i].x = mx+1-qr[i].x;}cdqdc(1,q);sort(qr+1,qr+q+1);for(int i=1; i<=q; i++) {qr[i].y = mx+1-qr[i].y;}cdqdc(1,q);sort(qr+1,qr+q+1);for(int i=1; i<=q; i++) {qr[i].x = mx+1-qr[i].x;}cdqdc(1,q);sort(qr+1,qr+q+1);for(int i=1; i<=q; i++){if(qr[i].opt==2) printf("%d\n",qr[i].ans);}return 0; }

總結(jié)

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