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题意:
给一段数字序列,求一段区间内未出现的最小自然数.
SOL:
框架显然用莫队.因为它兹瓷离线.
然而在统计上我打了线段树...用&维护的结点...400w的线段树...然后二分查找...炸的妥妥的...
然后发现所谓的"暴力"...直接开数组维护...因为指针具有一定的单调性,一次更改可以直接得到解,不用每次都查询...
woc真是...有时候数据结构用多了忽略了那些更简单更实用的方法...
Code
TLE的代码:
/*========================================================================== # Last modified: 2016-03-22 20:48 # Filename: 3339.cpp # Description: ==========================================================================*/ #define me AcrossTheSky #include <cstdio> #include <cmath> #include <ctime> #include <string> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #include <set> #include <map> #include <stack> #include <queue> #include <vector> #define lowbit(x) (x)&(-x) #define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++) #define FORP(i,a,b) for(int i=(a);i<=(b);i++) #define FORM(i,a,b) for(int i=(a);i>=(b);i--) #define ls(a,b) (((a)+(b)) << 1) #define rs(a,b) (((a)+(b)) >> 1) #define getlc(a) ch[(a)][0] #define getrc(a) ch[(a)][1] #define maxn 3600000 #define cap 201000 #define maxm 100000 #define pi 3.1415926535898 #define _e 2.718281828459 #define INF 1070000000 using namespace std; typedef long long ll; typedef unsigned long long ull; template<class T> inline void read(T& num) { bool start=false,neg=false; char c; num=0; while((c=getchar())!=EOF) { if(c==‘-‘) start=neg=true; else if(c>=‘0‘ && c<=‘9‘) { start=true; num=num*10+c-‘0‘; } else if(start) break; } if(neg) num=-num; } /*==================split line==================*/ struct Query{ int l,r,id,op; }q[cap]; int n,m; int v[maxn],a[cap],ans[cap]; int cmp(const Query &x,const Query &y){ if (x.op==y.op) return x.r<y.r; else return x.op<y.op; } void change(int node,int l,int r,int x,int d){ if (l==r) { v[node]+=d; return;} int mid=rs(l,r),lc=ls(node,0),rc=lc|1; if (x<=mid) change(lc,l,mid,x,d); else change(rc,mid+1,r,x,d); v[node]=(v[lc]?1:0)&(v[rc]?1:0); } int query(int node,int l,int r){ if (l==r) return l; int mid=rs(l,r),lc=ls(node,0),rc=lc|1; if (!v[lc]) query(lc,l,mid); else query(rc,mid+1,r); } void init(){ read(n); read(m); int sz=trunc(sqrt(n)); FORP(i,1,n) { read(a[i]); a[i]++;} FORP(i,1,m) { read(q[i].l); read(q[i].r); q[i].id=i; q[i].op=q[i].l/sz+(q[i].l%sz?1:0); } } int main(){ init(); sort(q+1,q+1+m,cmp); int L=q[1].l,R=q[1].r; FORP(i,L,R) change(1,1,cap,a[i],1); ans[q[1].id]=query(1,1,cap); FORP(i,2,m){ while (L<q[i].l) { change(1,1,cap,a[L],-1); L++;} while (L>q[i].l) { L--; change(1,1,cap,a[L],1);} while (q[i].r<R) { change(1,1,cap,a[R],-1); R--;} while (q[i].r>R) { R++; change(1,1,cap,a[R],1);} ans[q[i].id]=query(1,1,cap); } FORP(i,1,m) printf("%d\n",ans[i]-1); }
A掉的代码:
/*========================================================================== # Last modified: 2016-03-22 20:48 # Filename: 3339.cpp # Description: ==========================================================================*/ #define me AcrossTheSky #include <cstdio> #include <cmath> #include <ctime> #include <string> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #include <set> #include <map> #include <stack> #include <queue> #include <vector> #define lowbit(x) (x)&(-x) #define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++) #define FORP(i,a,b) for(int i=(a);i<=(b);i++) #define FORM(i,a,b) for(int i=(a);i>=(b);i--) #define ls(a,b) (((a)+(b)) << 1) #define rs(a,b) (((a)+(b)) >> 1) #define getlc(a) ch[(a)][0] #define getrc(a) ch[(a)][1] #define maxn 3600000 #define cap 201000 #define maxm 100000 #define pi 3.1415926535898 #define _e 2.718281828459 #define INF 1070000000 using namespace std; typedef long long ll; typedef unsigned long long ull; template<class T> inline void read(T& num) { bool start=false,neg=false; char c; num=0; while((c=getchar())!=EOF) { if(c==‘-‘) start=neg=true; else if(c>=‘0‘ && c<=‘9‘) { start=true; num=num*10+c-‘0‘; } else if(start) break; } if(neg) num=-num; } /*==================split line==================*/ struct Query{ int l,r,id,op; }q[cap]; int n,m; int cnt[maxn],a[cap],ans[cap]; int now=0; int cmp(const Query &x,const Query &y){ if (x.op==y.op) return x.r<y.r; else return x.op<y.op; } void change(int x,int d){ cnt[x]+=d; if (x<now){ if (cnt[x]==0) now=x; } else if (now==x) while(cnt[now]) now++; } void init(){ read(n); read(m); int sz=trunc(sqrt(n)); FORP(i,1,n) { read(a[i]);} FORP(i,1,m) { read(q[i].l); read(q[i].r); q[i].id=i; q[i].op=q[i].l/sz+(q[i].l%sz?1:0); } } int main(){ init(); sort(q+1,q+1+m,cmp); int L=q[1].l,R=q[1].r; FORP(i,L,R) change(a[i],1); ans[q[1].id]=now; FORP(i,2,m){ while (L<q[i].l) { change(a[L],-1); L++;} while (L>q[i].l) { L--; change(a[L],1);} while (q[i].r<R) { change(a[R],-1); R--;} while (q[i].r>R) { R++; change(a[R],1);} ans[q[i].id]=now; } FORP(i,1,m) printf("%d\n",ans[i]); }
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原文地址:http://www.cnblogs.com/YCuangWhen/p/5309525.html