bzoj 3339: Rmq Problem
3339: Rmq Problem
Time Limit: 20 Sec Memory Limit: 128 MBSubmit: 1270 Solved: 666
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Description
Input
Output
Sample Input
7 5
0 2 1 0 1 3 2
1 3
2 3
1 4
3 6
2 7
0 2 1 0 1 3 2
1 3
2 3
1 4
3 6
2 7
Sample Output
3
0
3
2
4
0
3
2
4
HINT
Source
嗯,莫队
懒得敲线段树,毕竟(明明)线段树比较短
#include<cmath> #include<cstdio> #include<algorithm> using namespace std; const int maxn = 200006; int sz,block[maxn]; struct Que{ int l,r,id; bool operator < (const Que&a)const { return block[l]==block[a.l] ? r<a.r : l<a.l; } }q[maxn]; int a[maxn],n,m,tmp,have[maxn],ans[maxn];int l=1,r=0; void add(int x) { ++have[x]; if(x==tmp||!x)while(have[tmp])++tmp; } void del(int x) { --have[x]; if(x < tmp&&!have[x])tmp=x; } void md(int &l,int &r,int ql,int qr,int id) { while(r < qr)add(a[++r]); while(r > qr)del(a[r--]); while(l < ql)del(a[l++]); while(l > ql)add(a[--l]); ans[id]=tmp; } int main() { scanf("%d%d",&n,&m); sz=sqrt(n); for(int i=1;i<=n;i++) scanf("%d",a+i),block[i]=(i-1)/sz+1; for(int i=1;i<=m;i++) scanf("%d%d",&q[i].l,&q[i].r),q[i].id=i; sort(q+1,q+m+1); for(int i=1;i<=m;++i) md(l,r,q[i].l,q[i].r,q[i].id); for(int i=1;i<=m;i++)printf("%d\n",ans[i]); return 0; }