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]);
}