本文主要是介绍Codeforces 220B - Little Elephant and Array 离线树状数组，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
This problem can be solve in simpler O(NsqrtN) solution, but I will describe O(NlogN) one.
We will solve this problem in offline. For each x (0 ≤ x < n) we should keep all the queries that end in x. Iterate that x from 0 to n - 1. Also we need to keep some array D such that for current x Dl + Dl + 1 + ... + Dx will be the answer for query [l;x]. To keep D correct, before the processing all queries that end in x, we need to update D. Let t be the current integer in A, i. e. Ax, and vector P be the list of indices of previous occurences of t (0-based numeration of vector). Then, if |P| ≥ t, you need to add 1 to DP[|P| - t], because this position is now the first (from right) that contains exactly t occurences of t. After that, if |P| > t, you need to subtract 2 from DP[|P| - t - 1], in order to close current interval and cancel previous. Finally, if |P| > t + 1, then you need additionally add 1 to DP[|P| - t - 2] to cancel previous close of the interval.
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)const int MAXN = 1e5+100;
int n,m;int a[MAXN],c[MAXN],ans[MAXN];
struct Query
{int l,r,id;bool operator < (const Query &t) const {return r<t.r;}
}q[MAXN];
inline int lowbit(int x){return x&(-x);}
void add(int i, int v)
{while(i<=n){c[i]+=v;i+=lowbit(i);}
}
int sum(int x)
{int ret=0;while(x>0){ret+=c[x];x-=lowbit(x);}return ret;
}int main()
{int sz;while(~scanf("%d%d",&n,&m)){vector<int>data[MAXN];CL(c,0);for(int i=1;i<=n;i++)scanf("%d",&a[i]);for(int i=1;i<=m;i++){scanf("%d%d",&q[i].l,&q[i].r);q[i].id=i;}sort(q+1,q+1+m);for(int i=1,k=1;i<=n;i++){if(a[i]<=n){data[a[i]].push_back(i);sz=data[a[i]].size();if(sz>=a[i]){add(data[a[i]][sz-a[i]],1);if(sz>a[i])add(data[a[i]][sz-a[i]-1],-2);if(sz>a[i]+1)add(data[a[i]][sz-a[i]-2],1);}}while(q[k].r==i && k<=m){ans[q[k].id]=sum(q[k].r)-sum(q[k].l-1);k++;}}for(int i=1;i<=m;i++)printf("%d\n",ans[i]);}return 0;
}
用于调试理解的及及加了注释的代码
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)const int MAXN = 1e5+100;
int n,m;int a[MAXN],c[MAXN],ans[MAXN];
struct Query
{int l,r,id;bool operator < (const Query &t) const {return r<t.r;}
}q[MAXN];
inline int lowbit(int x){return x&(-x);}
void add(int i, int v)
{while(i<=n){c[i]+=v;i+=lowbit(i);}
}
int sum(int x)
{int ret=0;while(x>0){ret+=c[x];x-=lowbit(x);}return ret;
}int main()
{int sz;while(~scanf("%d%d",&n,&m)){vector<int>data[MAXN];CL(c,0);for(int i=1;i<=n;i++)scanf("%d",&a[i]);for(int i=1;i<=m;i++){scanf("%d%d",&q[i].l,&q[i].r);q[i].id=i;}sort(q+1,q+1+m);for(int i=1,k=1;i<=n;i++){if(a[i]<=n){data[a[i]].push_back(i);sz=data[a[i]].size();if(sz>=a[i]){add(data[a[i]][sz-a[i]],1);//从右往左第a[i]次出现a[i]的位置+1if(sz>a[i])add(data[a[i]][sz-a[i]-1],-2);//从右往左第a[i]+1次出现a[i]的位置 -2,//因为当Sz==a[i]的时候，这个位置已经被加过1，此次读到i的时候，//从右往左第a[i]次出现a[i]的位置也被+1，//那么查询第a[i]+1次出现a[i]的位置到i，答案就是-2+1+1=0，//查询第a[i]次出现a[i]的位置到i,答案就是1if(sz>a[i]+1)add(data[a[i]][sz-a[i]-2],1);//从右往左第a[i]+2次出现a[i]的位置 +1,之前被+1-2，所以变成0//这三行代码维护出来，从当前的i往左数，第a[i]次出现a[i]的位置总是1//第a[i]+1次出现a[i]的位置总是-1，第a[i]+2及更多次的位置总是0,这样以i为右端点的区间的查询结果就都对了}}while(q[k].r==i && k<=m){/printf("#i=%d#\n",i);for(int j=0;j<=n;j++)printf("c[%d]=%d\n",j,c[j]);//ans[q[k].id]=sum(q[k].r)-sum(q[k].l-1);k++;}}for(int i=1;i<=m;i++)printf("%d\n",ans[i]);}return 0;
}
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