HDU 4638 group(离线+树状数组)

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-8
# define MOD 1000000007
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=100005;
//Code begin...

struct Q{int l, r, id;}q[N];
int a[N], to[N][2], 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=0;
    while (x) res+=tree[x], x-=lowbit(x);
    return res;
}
int main ()
{
    int T, m;
    T=Scan();
    while (T--) {
        mem(to,0); mem(vis,0); mem(tree,0);
        n=Scan(); m=Scan();
        FOR(i,1,n) {
            a[i]=Scan();
            if (vis[a[i]-1]) to[i][0]=vis[a[i]-1];
            if (vis[a[i]+1]) to[i][1]=vis[a[i]+1];
            vis[a[i]]=i;
        }
        FOR(i,1,m) q[i].l=Scan(), q[i].r=Scan(), q[i].id=i;
        sort(q+1,q+m+1,comp);
        int now=0;
        FOR(i,1,m) {
            while (now<q[i].r) {
                ++now;
                add(now,1);
                if (to[now][0]) add(to[now][0],-1);
                if (to[now][1]) add(to[now][1],-1);
            }
            ans[q[i].id]=sum(q[i].r)-sum(q[i].l-1);
        }
        FOR(i,1,m) Out(ans[i]), putchar('
');
    }
    return 0;
}