Codeforces 220B - Little Elephant and Array 离线树状数组

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+1                    if(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1                    if(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;}