BZOJ 3110 K大数查询 (树套树)

3110: [Zjoi2013]K大数查询

Time Limit: 20 Sec  Memory Limit: 512 MB
Submit: 9489  Solved: 2805
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Description

有N个位置，M个操作。操作有两种，每次操作如果是1 a b c的形式表示在第a个位置到第b个位置，每个位置加入一个数c
如果是2 a b c形式，表示询问从第a个位置到第b个位置，第C大的数是多少。

Input

第一行N，M
接下来M行，每行形如1 a b c或2 a b c

Output

输出每个询问的结果

Sample Input

2 5
1 1 2 1
1 1 2 2
2 1 1 2
2 1 1 1
2 1 2 3

Sample Output

1
2
1

HINT



【样例说明】

第一个操作 后位置 1 的数只有 1 ， 位置 2 的数也只有 1 。 第二个操作 后位置 1

的数有 1 、 2 ，位置 2 的数也有 1 、 2 。 第三次询问 位置 1 到位置 1 第 2 大的数 是

1 。 第四次询问 位置 1 到位置 1 第 1 大的数是 2 。 第五次询问 位置 1 到位置 2 第 3

大的数是 1 。‍


N,M<=50000,N,M<=50000

a<=b<=N

1操作中abs(c)<=N

2操作中c<=Maxlongint


Source

析：第一维用权值线段树，也就是来维护每个值，由于是有负数，所以，要多一开倍，询问的是第 k 大，可以把每个值c，都变成 n - c + 1，这样查询的时候就是第 k 小了，然后第二维用区间线段树，维护一个区间，用快速给一些区间加1操作，这样，就可以直接来查询某个区间的第 k 小了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-3;
const int maxn = 1e5 + 10;
const int maxm = (100000 << 8) + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

int root[maxn<<2], tot, lc[maxm], rc[maxm];
unsigned int sum[maxm], addv[maxm];

void push_up(int rt){
  sum[rt] = sum[lc[rt]] + sum[rc[rt]];
}

void push_down(int rt, int len){
  if(!lc[rt])  lc[rt] = ++tot;  // note
  if(!rc[rt])  rc[rt] = ++tot;  // note
  sum[lc[rt]] += (LL)(len - (len>>1)) * addv[rt];
  sum[rc[rt]] += (LL)(len >> 1) * addv[rt];
  addv[lc[rt]] += addv[rt];
  addv[rc[rt]] += addv[rt];
  addv[rt] = 0;
}

void update(int L, int R, int l, int r, int &rt){
  if(!rt)  rt = ++tot;
  if(L <= l && r <= R){
    sum[rt] += r - l + 1;
    ++addv[rt];
    return ;
  }
  if(addv[rt])  pd(rt, r - l + 1);
  int m = l + r >> 1;
  if(L <= m)  update(L, R, l, m, lc[rt]);
  if(R > m)   update(L, R, m+1, r, rc[rt]);
  pu(rt);
}

void update(int L, int R, int k){
  int l = 1, r = n<<1, rt = 1;
  while(l < r){
    int m = l + r >> 1;
    update(L, R, 1, n, root[rt]);
    if(k <= m)  rt <<= 1, r = m;
    else rt = rt<<1|1, l = m + 1;
  }
  update(L, R, 1, n, root[rt]);
}

unsigned int query(int L, int R, int l, int r, int rt){
  if(L <= l && r <= R)  return sum[rt];
  if(addv[rt])  pd(rt, r - l + 1);
  int m = l + r >> 1;
  unsigned int ans = 0;
  if(L <= m)  ans = query(L, R, l, m, lc[rt]);
  if(R > m)   ans += query(L, R, m+1, r, rc[rt]);
  return ans;
}

int query(int L, int R, int k){
  int l = 1, r = n<<1, rt = 1;
  unsigned int tmp;
  while(l < r){
    int m = l + r >> 1;
    if((tmp = query(L, R, 1, n, root[rt<<1])) >= k)  rt <<= 1, r = m;
    else rt = rt<<1|1, l = m + 1, k -= tmp;
  }
  return l;
}

int main(){
  scanf("%d %d", &n, &m);
  int op, l, r, k;
  while(m--){
    scanf("%d %d %d %d", &op, &l, &r, &k);
    if(op == 1)  update(l, r, n - k + 1);
    else printf("%d
", n - query(l, r, k) + 1);
  }
  return 0;
}

　整体二分，跑的真是快：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-3;
const int maxn = 1e5 + 40;
const int maxm = (100000 << 8) + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

int a[maxn], b[maxn], s[maxn];

void add(int x, int c){
  for(int i = x; i <= (n<<1); i += i&-i)  a[i] += c, b[i] += c * (x-1);
}

LL query(int x){
  int t0 = 0;  LL t1 = 0;
  for(int i = x; i; i -= -i&i)  t0 += a[i], t1 += b[i];
  return 1LL * x * t0 - t1;
}

struct Query{
  int ty, l, r, id, k;
};
Query q[maxn], q1[maxn], q2[maxn];
int ans[maxn];

void dfs(int fro, int rear, int l, int r){
  if(l > r || fro > rear)  return ;
  if(l == r){
    for(int i = fro; i <= rear; ++i)
      if(q[i].id != -1)  ans[q[i].id] = n - l + 1;
    return ;
  }
  int m = l + r >> 1;
  int lx = 0, rx = 0;
  for(int i = fro; i <= rear; ++i){
    if(q[i].ty == 1){
      if(q[i].k <= m){
        add(q[i].l, 1);
        add(q[i].r+1, -1);
        q1[lx++] = q[i];
      }
      else  q2[rx++] = q[i];
    }
    else{
      LL t = query(q[i].r) - query(q[i].l-1);
      if(t >= q[i].k)  q1[lx++] = q[i];
      else q[i].k -= t, q2[rx++] = q[i];
    }
  }
  for(int i = fro; i <= rear; ++i)
    if(q[i].ty == 1 && q[i].k <= m){
      add(q[i].l, -1);  add(q[i].r+1, 1);
    }
  for(int i = fro, j = 0; j < lx; ++j, ++i)  q[i] = q1[j];
  for(int i = lx+fro, j = 0; j < rx; ++j, ++i)  q[i] = q2[j];
  dfs(fro, fro + lx - 1, l, m);
  dfs(fro + lx, rear, m+1, r);
}

int main(){
  scanf("%d %d", &n, &m);
  int idx = 0;
  for(int i = 0; i < m; ++i){
    scanf("%d %d %d %d", &q[i].ty, &q[i].l, &q[i].r, &q[i].k);
    if(q[i].ty == 1){
      q[i].k = n - q[i].k + 1;
      q[i].id = -1;
    }
    else q[i].id = idx++;
  }
  dfs(0, m-1, 0, (n<<1)+2);
  for(int i = 0; i < idx; ++i)  printf("%d
", ans[i]);
  return 0;
}

　　

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原文地址：https://www.cnblogs.com/dwtfukgv/p/7763464.html