HDU 4638 group(离线+树状数组)

时间：2023-01-26 00:31:11

此题实质上是询问每段区间[l,r]内的数字能组成多少段连续的数字。

不大好用线段树合并区间来写。

考虑离线，类似于HH的项链的话，对询问按右端点排序，对于右端点的每一次右移，则对于询问，实际上是求的后缀区间。

如果前面已经出现了a[r]-1的话，那么它们能组成连续的区间，a[r]+1同理。

剩下的模仿HH的项链即可。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int res=, flag=;

    char ch;

    if((ch=getchar())=='-') flag=;

    else if(ch>=''&&ch<='') res=ch-'';

    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');

    return flag?-res:res;

}

void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

struct Q{int l, r, id;}q[N];

int a[N], to[N][], vis[N], tree[N], ans[N], n;

bool comp(Q a, Q b){return a.r<b.r;}

void add(int x, int val){while (x<=n) tree[x]+=val, x+=lowbit(x);}

int sum(int x)

{

    int res=;

    while (x) res+=tree[x], x-=lowbit(x);

    return res;

}

int main ()

{

    int T, m;

    T=Scan();

    while (T--) {

        mem(to,); mem(vis,); mem(tree,);

        n=Scan(); m=Scan();

        FOR(i,,n) {

            a[i]=Scan();

            if (vis[a[i]-]) to[i][]=vis[a[i]-];

            if (vis[a[i]+]) to[i][]=vis[a[i]+];

            vis[a[i]]=i;

        }

        FOR(i,,m) q[i].l=Scan(), q[i].r=Scan(), q[i].id=i;

        sort(q+,q+m+,comp);

        int now=;

        FOR(i,,m) {

            while (now<q[i].r) {

                ++now;

                add(now,);

                if (to[now][]) add(to[now][],-);

                if (to[now][]) add(to[now][],-);

            }

            ans[q[i].id]=sum(q[i].r)-sum(q[i].l-);

        }

        FOR(i,,m) Out(ans[i]), putchar('\n');

    }

    return ;

}

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