给出B,I,R,对于Pi,若存在Pj满足 Bi < Bj && Ii < Ij && Ri < Rj ,则称Pi为 probable self-murderers。问存在多少个Pi。
首先对其升序排序,优先级为B > I > R。
然后发现对于i < j ,必有Bi <= Bj,所以这一层基本可以忽略。
然后由于数据范围较大,对 I 进行离散,建立线段树记录大于I的区间内最大的R是多少,当然此时要从后往前扫描,边更新边计数。
#include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <queue> #include <cmath> #include <stack> #include <map> #include <ctime> #include <iomanip> #pragma comment(linker, "/STACK:1024000000"); #define EPS (1e-6) #define LL long long #define ULL unsigned long long #define _LL __int64 #define INF 0x3f3f3f3f #define Mod 1000000007 using namespace std; struct N { int b,s,r; } sta[500010]; bool cmp(N n1,N n2) { if(n1.b != n2.b) return n1.b < n2.b; if(n1.s != n2.s) return n1.s < n2.s; return n1.r < n2.r; } int num[500010]; int DelSame(int *num,int Top) { int i,j; for(i = 2,j = 1; i <= Top; ++i) if(num[i] != num[j]) num[++j] = num[i]; return j; } int BS(int *num,int l,int r,int x) { int mid; while(l <= r) { mid = (l+r)>>1; if(num[mid] == x) return mid; if(num[mid] < x) l = mid+1; else r = mid-1; } return -1; } int st[2000010]; int Updata(int site,int l,int r,int x,int d) { if(l == r && r == x) return st[site] = max(st[site],d); int mid = (l+r)>>1; if(x <= mid) return st[site] = max(st[site],Updata(site<<1,l,mid,x,d)); return st[site] = max(st[site],Updata(site<<1|1,mid+1,r,x,d)); } int Query(int site,int L,int R,int l,int r) { if(L == l && R == r) return st[site]; int mid = (L+R)>>1; if(r <= mid) return Query(site<<1,L,mid,l,r); if(mid < l) return Query(site<<1|1,mid+1,R,l,r); return max( Query(site<<1,L,mid,l,mid), Query(site<<1|1,mid+1,R,mid+1,r) ); } int main() { int n,i,j; scanf("%d",&n); for(i = 0; i < n; ++i) scanf("%d",&sta[i].b); for(i = 0; i < n; ++i) scanf("%d",&sta[i].r); for(i = 0; i < n; ++i) scanf("%d",&sta[i].s); sort(sta,sta+n,cmp); int Top = 0; for(i = 0; i < n; ++i) num[++Top] = sta[i].r; sort(num+1,num+Top+1); Top = DelSame(num,Top); // for(i = 1;i <= Top; ++i) // printf("%d %d\n",i,num[i]); // for(i = 0; i < n; ++i) // printf("i = %d %d %d %d\n",i,sta[i].b,sta[i].r,sta[i].s); memset(st,0,sizeof(st)); int site,ans = 0,tmp; int R = n-1,val = sta[n-1].b; for(i = n-1; i >= 0; --i) { if(sta[i].b == val) continue; for(j = i+1; j <= R; ++j) { site = BS(num,1,Top,sta[j].r); if(site != Top && (tmp = Query(1,1,Top,site+1,Top)) > sta[j].s) ans++; } for(j = i+1; j <= R; ++j) { site = BS(num,1,Top,sta[j].r); Updata(1,1,Top,site,sta[j].s); } R = i; val = sta[i].b; } for(j = i+1; j <= R; ++j) { site = BS(num,1,Top,sta[j].r); if(site != Top && (tmp = Query(1,1,Top,site+1,Top)) > sta[j].s) ans++; } printf("%d\n",ans); return 0; }
时间: 2024-10-14 15:16:22