划分树几道题目

主要参考kuangbin博客

个人感觉划分树是基于快速排序的分治的思想，是把快速排序每个过程记录下来而已（借助快速排序也是可以快速求整个区间第k值的）

对于每个区间，找中间值比较分成两个子区间，再递归处理

划分树Hdu4251（区间第k值）

typedef long long LL;typedef unsigned long long ULL;typedef vector <int> VI;const int INF = 1000000000;#define lson l, m, dep + 1#define rson m + 1, r, dep + 1const int maxn = 100010;const int deep = 30; int t[deep][maxn];int st[maxn];int toleft[deep][maxn]; void build(int l, int r, int dep){   int i;   if (l == r) return;   int m = (r + l) >> 1;   int same = m - l + 1;//表示等于中间值而且被分入左边的个数   for (i = l; i <= r; i++)       if (t[dep][i] < st[m])           same--;   int lpos = l;   int rpos = m + 1;   for (i = l; i <= r; i++)    {       if (t[dep][i] < st[m])//比中间的数小，分入左边           t[dep + 1][lpos++] = t[dep][i];       else if (t[dep][i] == st[m] && same > 0)       {           t[dep + 1][lpos++] = t[dep][i];           same--;       }       else// if (t[dep][i] > st[m]) //比中间值大分入右边            t[dep + 1][rpos++] = t[dep][i];       toleft[dep][i] = toleft[dep][l - 1] + lpos - l;//从1到i放左边的个数    }   build(lson);   build(rson);} //查询区间第k大的数，[l,r]是大区间，[L,R]是要查询的小区间int query(int L, int R, int k,int l, int r,int dep){   if (l == r) return t[dep][l];///!!!!!!!!!!!!!!!!!!!!!!!!!!!!1   int m = (l + r) >> 1;   int cnt = toleft[dep][R] - toleft[dep][L - 1];//[L,R]中位于左边的个数   if (cnt >= k)    {       //l+要查询的区间前被放在左边的个数       int newl = l + toleft[dep][L - 1] - toleft[dep][l - 1];       //左端点加上查询区间会被放在左边的个数       int newr = newl + cnt - 1;       return query(newl, newr, k, lson);///!!!!!newl, newr    }   else    {       int newr = R + toleft[dep][r] - toleft[dep][R];       int newl = newr - (R - L - cnt);       return query(newl, newr, k - cnt, rson);///!!!!!!    }} int main(){   int n, m;   int a, b;   int p = 1;   while (cin >> n)    {       CLR(t, 0);///       FE(i, 1, n)///       {           RI(t[0][i]);           st[i] = t[0][i];       }       sort(st + 1, st + n + 1);///       build(1, n, 0);///       printf("Case %d:\n",p++);       cin >> m;       REP(i, m)       {           RII(a, b);           printf("%d\n",query(a, b, (b - a) / 2 + 1, 1, n, 0));       }    }}

划分树hdu 3473 (Minimum Sum )增加了一个lsum【】函数

typedef long long LL;typedef unsigned long long ULL;typedef vector <int> VI;const int INF = 1000000000;const int maxn = 100010;const int deep = 20; #define lson l, m, dep + 1#define rson m + 1, r, dep + 1 int t[deep][maxn];int toleft[deep][maxn];int st[maxn];LL lsum[deep][maxn];//LL sum[maxn];LL ans; void build(int l, int r, int dep){   if (l == r) {       lsum[dep][l] = t[dep][l];       return ;    }   int m = (l + r) >> 1;   int same = m - l + 1;   FE(i, l, r)    {       if (t[dep][i] < st[m])           same--;       lsum[dep][i] = t[dep][i];       if (i != l) lsum[dep][i] += lsum[dep][i - 1];    }   int lpos = l;   int rpos = m + 1;   FE(i, l, r)    {       if (t[dep][i] < st[m])           t[dep + 1][lpos++] = t[dep][i];       else if (t[dep][i] == st[m] && same > 0)       {           t[dep + 1][lpos++] = t[dep][i];           same--;       }       else           t[dep + 1][rpos++] = t[dep][i];       toleft[dep][i] = toleft[dep][l - 1] + lpos - l;///    }   build(lson);   build(rson);} int query(int L,int R, int k, int l, int r,int dep){   if (L == R ) return t[dep][L];   int cnt = toleft[dep][R] - toleft[dep][L - 1];   int m = (l + r) >> 1;    int ln1 = toleft[dep][L - 1] - toleft[dep][l - 1];   int rn1 = L - l - ln1;    int ln2 = cnt;   int rn2 = R - L + 1 - cnt;    if (cnt >= k)    {       if (rn2 > 0)       {           if (rn1 > 0)                ans += lsum[dep + 1][m + rn1 +rn2] - lsum[dep + 1][m + rn1];///@@@@///dep + 1           else ans += lsum[dep + 1][m + rn2];       }        int newl = l + ln1;       int newr = newl + cnt - 1;       return query(newl, newr, k, lson);    }   else    {       if (ln2 > 0)       {           if (ln1 > 0)                ans -= lsum[dep + 1][l - 1 +ln1 + ln2] - lsum[dep + 1][l - 1 + ln1];           else ans -= lsum[dep + 1][l - 1 + ln2];       }        int newr = R + toleft[dep][r] - toleft[dep][R];       int newl = newr - (R - L - cnt);       return query(newl, newr, k - cnt, rson);    } } int T;int n, m;int main(){   int x, y;   cin >> T;   int p = 1;   while (T--)    {       RI(n);//       CLR(t, 0,);/////       CLR(toleft, 0);///       FE(i, 1, n)       {           RI(t[0][i]);           st[i] = t[0][i];       }       sort(st + 1, st + n + 1);       build(1, n, 0);///       printf("Case #%d:\n",p++);       RI(m);       REP(i, m)       {           RII(x, y);           x++;           y++;           ans = 0;           LL midval = query(x, y, (y - x) / 2 + 1, 1, n, 0);///0           if ((y - x + 1) % 2 == 0) ans -= midval;           cout << ans << endl;       }       printf("\n");    }} 

又一解法主要是lsum的处理不同

const int MAXN = 100010;const int MOD = 1000000;#define lson l, m, dep + 1#define rson m + 1, r, dep + 1int t[20][MAXN];int st[MAXN];int tol[20][MAXN];LL lsum[20][MAXN];LL ans;int n, m;void build(int l, int r, int dep){    if (l == r)    {        return ;    }    int m = (l + r) >> 1;    int same = (m - l + 1);    FE(i, l, r) if (st[m] > t[dep][i]) same--;    int lpos = l;    int rpos = m + 1;    FE(i, l, r)    {        if (t[dep][i] < st[m]) t[dep + 1][lpos++] = t[dep][i];        else if (t[dep][i] == st[m] && same > 0) t[dep + 1][lpos++] = t[dep][i], same--;        else t[dep + 1][rpos++] = t[dep][i];        tol[dep][i] = tol[dep][l - 1] + lpos - l;    }    FE(i, l, r) lsum[dep + 1][i] = lsum[dep + 1][i - 1] + t[dep + 1][i];    build(lson);    build(rson);}LL query(int L, int R, int k, int l, int r, int dep){    if (L == R) return t[dep][L];    int m = (r + l) >> 1;    int cnt = tol[dep][R] - tol[dep][L - 1];    int ln1 = tol[dep][L - 1] - tol[dep][l - 1];    int rn1 = L - l - ln1;    int ln2 = cnt;    int rn2 = R - L - cnt + 1;    if (cnt >= k)//toleft    {        if (rn2 > 0) ans += lsum[dep + 1][m + rn2 + rn1] - lsum[dep + 1][m + rn1];        int newl = l + tol[dep][L - 1] - tol[dep][l - 1];        int newr = newl + cnt - 1;        query(newl, newr, k, lson);    }    else//toright    {        if (ln2 > 0) ans -= lsum[dep + 1][l + ln2 + ln1 - 1] - lsum[dep + 1][l + ln1 - 1];        int newr = R + tol[dep][r] - tol[dep][R];        int newl = newr - (R - L - cnt);        query(newl, newr, k - cnt, rson);    }}int main(){    int T;    RI(T);    int x, y;    int ncase = 1;    while (T--)    {        RI(n);        CLR(t, 0);        CLR(lsum, 0);        CLR(tol, 0);        FE(i, 1, n)        {            RI(st[i]);            t[0][i] = lsum[0][i] = st[i];        }        sort(st + 1, st + n + 1);        FE(i, 1, n) lsum[0][i] += lsum[0][i - 1];        build(1, n, 0);        RI(m);                printf("Case #%d:\n", ncase++);        while (m--)        {            RII(x, y);            x++;            y++;            ans = 0;            int mid = query(x, y, (y - x) / 2 + 1, 1, n, 0);            if ((y - x + 1) % 2 == 0) ans -= mid;            cout << ans << endl;        }        cout << endl;    }}

划分树hdu 4417SuperMario （求给定数在给定区间的中比它小的数的个数，或大小位置）（求逆序数）(通过修改query)（也可以二分答案！！！！！！！！！！！）

typedef long long LL;typedef unsigned long long ULL;typedef vector <int> VI;const int INF = 1000000000;const int maxn = 100010;const int deep = 20; #define lson l, m, dep + 1#define rson m + 1, r, dep + 1 int t[deep][maxn];int tol[deep][maxn];int st[maxn];int n;void build(int l, int r, int dep){   if (l == r) return;   int m = (l + r) >> 1;   int same = m - l + 1;   FE(i, l, r)       if (t[dep][i] < st[m]) same--;   int lpos = l;   int rpos = m + 1;   FE(i, l, r)    {       if (t[dep][i] < st[m]) t[dep + 1][lpos++] = t[dep][i];       else if (t[dep][i] == st[m] && same > 0) { t[dep + 1][lpos++]=  t[dep][i]; same--; }       else t[dep + 1][rpos++] = t[dep][i];       tol[dep][i] = tol[dep][l - 1] + lpos - l;    }   build(lson);   build(rson);} int ans;int h;void query(int L, int R, int l, int r, intdep){   if (L == R)    {       if (t[dep][L] <= h)           ans++;       return ;    }   int m = (l + r) >> 1;   int cnt = tol[dep][R] - tol[dep][L - 1]; //   int ln1 = tol[dep][L - 1] - tol[dep][l - 1];//   int rn1 = L - l - ln1;////   int ln2 = cnt;//   int rn2 = R - L - cnt + 1;////   int lr = t[dep + 1][l - 1 + ln1 + ln2];//   int rl = t[dep + 1][m + rn1 + 1];    if (st[m] > h)    {       int newl = l + tol[dep][L - 1] - tol[dep][l - 1];       int newr = newl + cnt - 1;       if (newl <= newr)///?！！！！！！！！！！！       query(newl, newr, lson);    }    else    {       ans += cnt;       int newr = R + tol[dep][r] - tol[dep][R];       int newl = newr - (R - L - cnt);       if (newl <= newr)///?！！！！！！！！！！！       query(newl, newr, rson);    }} int T; int main(){   int x, y, z;   int T;   int p = 1;   RI(T);   while (T--)    {       //初始化       printf("Case %d:\n",p++);       CLR(t, 0);       CLR(tol, 0);       int m;       RII(n, m);       FE(i, 1, n)///       {           RI(t[0][i]);           st[i] = t[0][i];       }       sort(st + 1, st + n + 1);       build(1, n, 0);        while (m--)       {           RIII(x, y, z);           h = z;           ans = 0;           query(++x, ++y, 1, n, 0);           cout << ans << endl;       }    }}