问题 D: A Sequence Game
时间限制: 1 Sec 内存限制: 128 MB
提交: 148 解决: 42
[提交] [状态] [讨论版] [命题人:admin]
题目描述
One day, WNJXYK found a very hard problem on an Online Judge. This problem is so hard that he had been thinking about the solutions for a couple of days. And then he had a surprise that he misunderstood that problem and easily figured out a solution using segment tree. Now he still wonders that solution for the misread problem.
There is a sequence with N positive integers A1,A2,…,An and M queries. Each query will give you an interval [L,R] and require an answer with YES / NO indicates that whether the numbers in this interval are continuous in its integer range.
Let us assume that the maximal number in an interval is mx and the minimal number is mi. The numbers in this interval are continuous in its integer range means that each number from mi to mx appears at least once in this interval.
输入
The input starts with one line contains exactly one positive integer T which is the number of test cases. And then there are T cases follow.
The first line contains two positive integers n,m which has been explained above.
The second line contains n positive integers A1,A2,…,An.
Then there will be m lines followed. Each line contains to positive numbers Li,Ri indicating that the i th query’s interval is [Li,Ri].
输出
For each test case, output m line.
Each of following m lines contains a single string “YES”/ “NO” which is the answer you have got.
样例输入
2 3 3 3 1 2 2 3 1 3 1 2 5 3 1 2 2 4 5 1 5 1 3 3 3
样例输出
YES YES NO NO YES YES
提示
T=5
1≤n≤100000
1≤Ai≤10^9
1≤m≤100000
The input file is very large, so you are recommend to use scanf() and printf() for IO.
题意:给你一个长度为n的序列,有m次询问,每次问询一个区间中是否区间中最大的数到最小的数都出现了至少一次
题解:明显可以用莫队算法对于区间查询进行排序然后分块处理,每次查询区间的最大值和最小值然后莫队处理区间中不同的数字的个数,若最大值减去最小值+1<=区间中不同数字的个数,则输出YES否则输出NO,区间RMQ可以用ST表进行优化,复杂度O(n*sqrt(n)),一开始我用map标记区间中出现的数字的个数总是运行错误,将数据离散化再用数组标记就可以了;
#include<bits/stdc++.h>
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define SI(i) scanf("%lld",&i)
#define PI(i) printf("%lld\n",i)
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int MAX=2e5+5;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
int dir[9][2]={0,1,0,-1,1,0,-1,0, -1,-1,-1,1,1,-1,1,1};
template<class T>bool gmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>bool gmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>void gmod(T &a,T b){a=((a+b)%mod+mod)%mod;}
typedef pair<ll,ll> PII;
ll n,cnt,st[MAX][25][2],powll[50],lg[MAX],ans[MAX],vis[MAX],a[MAX],b[MAX];
int del(ll x)
{
x=st[x][0][0];
if(vis[x]==1) cnt--;
vis[x]--;
}
int add(ll x)
{
x=st[x][0][0];
if(vis[x]==0) cnt++;
vis[x]++;
}
struct node
{
ll l,r,id;
}p[MAX];
int block;
int cmp(node a,node b)
{
if(!block) block++;
if( (a.l/block) == (b.l/block) )
return a.r<b.r;
return a.l<b.l;
}
int main()
{
powll[0]=1;
for(int i=1;i<=25;i++)
powll[i]=(powll[i-1]<<1);
lg[0]=-1;
int T;
scanf("%d",&T);
while(T--)
{
memset(vis,0,sizeof(vis));
ll m;
scanf("%lld%lld",&n,&m);
block=(int)(sqrt(n*1.0)+0.5);
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]),lg[i]=lg[i>>1]+1,b[i]=a[i];
sort(b+1,b+1+n);
for(int i=1;i<=n;i++)
{
st[i][0][0]=lower_bound(b+1,b+1+n,a[i])-b-1;
st[i][0][1]=st[i][0][0];
}
for(int j=1;(1<<j)<=n;j++)
for(int i=1;i<=n;i++)
{
st[i][j][0]=max(st[i][j-1][0],st[i+(1<<j-1)][j-1][0]);
st[i][j][1]=min(st[i][j-1][1],st[i+(1<<j-1)][j-1][1]);
}
for(int i=1;i<=m;i++)
{
scanf("%lld%lld",&p[i].l,&p[i].r);
p[i].id=i;
}
sort(p+1,p+1+m,cmp);
ll L=1,R=0;cnt=0;
for(int i=1;i<=m;i++)
{
ll l=p[i].l;
ll r=p[i].r;
while(L<l) del(L++);
while(L>l) add(--L);
while(R>r) del(R--);
while(R<r) add(++R);
ll d=lg[r-l+1];
r=r-powll[d]+1;
ll a=min(st[l][d][1],st[r][d][1]);
ll b=max(st[l][d][0],st[r][d][0]);
if(cnt>=b-a+1)
ans[p[i].id]=1;
else
ans[p[i].id]=0;
}
for(int i=1;i<=m;i++)
if(ans[i])
printf("YES\n");
else
printf("NO\n");
}
return 0;
}