https://www.cnblogs.com/tsw123/p/4374728.html

点修改

Update(x,v):  把A[x]修改为v

Query(L,R): 计算区间[L,R] 最小值.

// Dynamic RMQ
// Rujia Liu
// 输入格式:
// n m    数组范围是a[1]~a[n],初始化为0。操作有m个
// 1 p v  表示设a[p]=v
// 2 L R  查询a[L]~a[R]的min
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int INF = 1000000000;
const int maxnode = 1<<17;

int op, qL, qR, p, v;  //qL和qR为全局变量,询问区间[qL,qR];

struct IntervalTree {
  int minv[maxnode];

  void update(int o, int L, int R) {
    int M = L + (R-L)/2;
    if(L == R) minv[o] = v; // 叶结点,直接更新minv
    else {
      // 先递归更新左子树或右子树
      if(p <= M) update(o*2, L, M); else update(o*2+1, M+1, R);
      // 然后计算本结点的minv
      minv[o] = min(minv[o*2], minv[o*2+1]);
    }
  }

  int query(int o, int L, int R) {
    int M = L + (R-L)/2, ans = INF;
    if(qL <= L && R <= qR) return minv[o]; // 当前结点完全包含在查询区间内
    if(qL <= M) ans = min(ans, query(o*2, L, M)); // 往左走
    if(M < qR) ans = min(ans, query(o*2+1, M+1, R)); // 往右走
    return ans;
  }
};


IntervalTree tree;

int main() {
  int n, m;
  while(scanf("%d%d", &n, &m) == 2) {
    memset(&tree, 0, sizeof(tree));
    while(m--) {
      scanf("%d", &op);
      if(op == 1) {
        scanf("%d%d", &p, &v);
        tree.update(1, 1, n);  // 修改树节点,或者是建树的过程
      } else {
        scanf("%d%d", &qL, &qR);  //修改询问区间
        printf("%d\n", tree.query(1, 1, n));
      }
    }
  }
  return 0;
}
View Code

区间修改

一段区间加上一个数求最大值、最小值、和

// Fast Sequence Operations I
// Rujia Liu
// 输入格式:
// n m     数组范围是a[1]~a[n],初始化为0。操作有m个
// 1 L R v 表示设a[L]+=v, a[L+1]+v, ..., a[R]+=v
// 2 L R   查询a[L]~a[R]的sum, min和max
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxnode = 1<<17;

int _sum, _min, _max, op, qL, qR, v; //<span style="color:#ff0000;">_sum为全局变量</span>

struct IntervalTree {
  int sumv[maxnode], minv[maxnode], maxv[maxnode], addv[maxnode];

  // 维护信息
  void maintain(int o, int L, int R) {
    int lc = o*2, rc = o*2+1;
    sumv[o] = minv[o] = maxv[o] = 0;
    if(R > L) {
      sumv[o] = sumv[lc] + sumv[rc];
      minv[o] = min(minv[lc], minv[rc]);
      maxv[o] = max(maxv[lc], maxv[rc]);
    }
    if(addv[o]) { minv[o] += addv[o]; maxv[o] += addv[o]; sumv[o] += addv[o] * (R-L+1); }
  }

  void update(int o, int L, int R) {
    int lc = o*2, rc = o*2+1;
    if(qL <= L && qR >= R) { // 递归边界
      addv[o] += v; // 累加边界的add值
    } else {
      int M = L + (R-L)/2;
      if(qL <= M) update(lc, L, M);
      if(qR > M) update(rc, M+1, R);
    }
    maintain(o, L, R); // 递归结束前重新计算本结点的附加信息
  }

  void query(int o, int L, int R, int add) {
    if(qL <= L && qR >= R) { // 递归边界:用边界区间的附加信息更新答案
      _sum += sumv[o] + add * (R-L+1);
      _min = min(_min, minv[o] + add);
      _max = max(_max, maxv[o] + add);
    } else { // 递归统计,累加参数add
      int M = L + (R-L)/2;
      if(qL <= M) query(o*2, L, M, add + addv[o]);
      if(qR > M) query(o*2+1, M+1, R, add + addv[o]);
    }
  }
};

const int INF = 1000000000;

IntervalTree tree;

int main() {
  int n, m;
  while(scanf("%d%d", &n, &m) == 2) {
    memset(&tree, 0, sizeof(tree));
    while(m--) {
      scanf("%d%d%d", &op, &qL, &qR);
      if(op == 1) {
        scanf("%d", &v);
        tree.update(1, 1, n);
      } else {
        _sum = 0; _min = INF; _max = -INF;
        tree.query(1, 1, n, 0);
        printf("%d %d %d\n", _sum, _min, _max);
      }
    }
  }
  return 0;
}
View Code

修改区间的值

求区间的和,最大值,最小值

// Fast Sequence Operations II
// Rujia Liu
// 输入格式:
// n m     数组范围是a[1]~a[n],初始化为0。操作有m个
// 1 L R v 表示设a[L]=a[L+1]=...=a[R] = v。其中v > 0
// 2 L R  查询a[L]~a[R]的sum, min和max
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxnode = 1<<17;

int _sum, _min, _max, op, qL, qR, v;

struct IntervalTree {
  int sumv[maxnode], minv[maxnode], maxv[maxnode], setv[maxnode];

  // 维护信息
  void maintain(int o, int L, int R) {
    int lc = o*2, rc = o*2+1;
    if(R > L) {
      sumv[o] = sumv[lc] + sumv[rc];
      minv[o] = min(minv[lc], minv[rc]);
      maxv[o] = max(maxv[lc], maxv[rc]);
    }
    if(setv[o] >= 0) { minv[o] = maxv[o] = setv[o]; sumv[o] = setv[o] * (R-L+1); }
  }

  // 标记传递
  void pushdown(int o) {
    int lc = o*2, rc = o*2+1;
    if(setv[o] >= 0) { //本结点有标记才传递。注意本题中set值非负,所以-1代表没有标记
      setv[lc] = setv[rc] = setv[o];
      setv[o] = -1; // 清除本结点标记
    }
  }

  void update(int o, int L, int R) {
    int lc = o*2, rc = o*2+1;
    if(qL <= L && qR >= R) { // 标记修改
      setv[o] = v;
    } else {
      pushdown(o);
      int M = L + (R-L)/2;
      if(qL <= M) update(lc, L, M); else maintain(lc, L, M);
      if(qR > M) update(rc, M+1, R); else maintain(rc, M+1, R);
    }
    maintain(o, L, R);
  }

  void query(int o, int L, int R) {
    if(setv[o] >= 0) { // 递归边界1:有set标记
      _sum += setv[o] * (min(R,qR)-max(L,qL)+1);
      _min = min(_min, setv[o]);
      _max = max(_max, setv[o]);
    } else if(qL <= L && qR >= R) { // 递归边界2:边界区间
      _sum += sumv[o]; // 此边界区间没有被任何set操作影响
      _min = min(_min, minv[o]);
      _max = max(_max, maxv[o]);
    } else { // 递归统计
      int M = L + (R-L)/2;
      if(qL <= M) query(o*2, L, M);
      if(qR > M) query(o*2+1, M+1, R);
    }
  }
};

const int INF = 1000000000;

IntervalTree tree;

int main() {
  int n, m;
  while(scanf("%d%d", &n, &m) == 2) {
    memset(&tree, 0, sizeof(tree));
    memset(tree.setv, -1, sizeof(tree.setv));
    tree.setv[1] = 0;
    while(m--) {
      scanf("%d%d%d", &op, &qL, &qR);
      if(op == 1) {
        scanf("%d", &v);
        tree.update(1, 1, n);
      } else {
        _sum = 0; _min = INF; _max = -INF;
        tree.query(1, 1, n);
        printf("%d %d %d\n", _sum, _min, _max);
      }
    }
  }
  return 0;
}
View Code

最后附上蒟蒻自己写的板子,可能有不对的地方,欢迎指正。

#include<bits/stdc++.h>
#define _for(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
typedef long long ll;
const int mod =1e6+7;
double esp=1e-6;
int INF =0x3f3f3f3f;
const int inf = 1<<28;
const int MAXN=1e5+5;
struct ST
{
    int num,_max,_min,_sum,l,r,lz;
    ST()
    {
        lz=0;
    }
}tree[MAXN*4];
void build(int l,int r,int id)
{
    tree[id].l=l;
    tree[id].r=r;
    tree[id].lz=0;
    if(l==r)
    {
        scanf("%d",&tree[id].num);
        tree[id]._sum=tree[id].num;
        tree[id]._max=tree[id].num;
        tree[id]._min=tree[id].num;
        //printf("%d:%d\n",id,tree[id].num);
        return ;
    }
    int mid=(l+r)>>1;
    build(l,mid,id*2);
    build(mid+1,r,id*2+1);
    tree[id]._sum=tree[id*2]._sum+tree[id*2+1]._sum;
    //printf("%d:%d\n",id,tree[id].)
    tree[id]._max=max(tree[id*2]._max,tree[id*2+1]._max);
    tree[id]._min=min(tree[id*2]._min,tree[id*2+1]._min);
    //printf("%d:%d %d\n",id,tree[id]._max,tree[id]._min);
}
void update(int id,int k,int num)//单点修改
{
    if(tree[id].l==tree[id].r)
    {
        //printf("**%d\n",k);
        tree[id].num=num;
        return ;
    }
    if(k<=tree[id*2].r)update(id*2,k,num);
    else update(id*2+1,k,num);
    tree[id]._sum=tree[id*2]._sum+tree[id*2+1]._sum;
    tree[id]._max=max(tree[id*2]._max,tree[id*2+1]._max);
    tree[id]._min=min(tree[id*2]._min,tree[id*2+1]._min);
}
int Seach(int id,int k)//单点查询
{
    if(tree[id].l==tree[id].r)
        return tree[id].num;
    if(k<=tree[2*id].r)
        Seach(id*2,k);
    else
        Seach(id*2+1,k);

}
void push_down(int id)//下放lz标记
{
    if(tree[id].lz!=0)
    {
        tree[id*2].lz+=tree[id].lz;
        tree[id*2+1].lz+=tree[id].lz;
        int mid=(tree[id].l+tree[id].r)>>1;
        tree[id*2]._sum+=tree[id*2].lz*(mid-tree[id].l+1);
        tree[id*2+1]._sum+=tree[id*2].lz*(tree[id].r-mid);
        tree[id*2]._max+=tree[id].lz;
        tree[id*2+1]._max+=tree[id].lz;
        tree[id*2]._min+=tree[id].lz;
        tree[id*2+1]._min+=tree[id].lz;
        tree[id].lz=0;
    }
    return ;
}
void updata_Qu(int l,int r,int id,int num)//l-r区间加上num
{
    if(tree[id].r<=r&&tree[id].l>=l)
    {
        tree[id]._sum+=num*(tree[id].r-tree[id].l+1);
        tree[id]._max=max(tree[id].r+num,tree[id].l+num);
        tree[id]._min=min(tree[id].r+num,tree[id].l+num);
        tree[id].lz+=num;
        return ;
    }
    push_down(id);
    if(tree[id*2].r>=l)
        updata_Qu(l,r,id*2,num);
    if(tree[id*2+1].l<=r)
        updata_Qu(l,r,id*2+1,num);
    tree[id]._sum=tree[id*2]._sum+tree[id*2+1]._sum;
    tree[id]._max=max(tree[id*2]._max,tree[id*2+1]._max);
    tree[id]._min=min(tree[id*2]._min,tree[id*2+1]._min);
    return ;
}
int Ma,Mi;
int Seach_Qu(int l,int r,int id)//区间查找
{
    if(tree[id].l>=l&&tree[id].r<=r)
    {
        Ma=max(tree[id]._max,Ma);
        Mi=min(tree[id]._min,Mi);
        return tree[id]._sum;
    }
    push_down(id);
    int s=0;
    if(tree[2*id].r>=l){s+=Seach_Qu(l,r,id*2);}//Mi=min(Mi,tree[id*2]._min);Ma=max(Ma,tree[id*2]._max);}
    if(tree[2*id+1].l<=r){s+=Seach_Qu(l,r,id*2+1);}//Mi=min(Mi,tree[id*2+1]._min);Ma=max(Ma,tree[id*2+1]._max);}
    return s;
}
int main()
{
    int n;
    scanf("%d",&n);
    build(1,n,1);
    updata_Qu(1,n,1,4);//1-n区间加上4
    Ma=-INF,Mi=INF;
    int mm=Seach_Qu(2,5,1);//查找[2-5]修改完后的和,最大值,最小值
    printf("%d %d %d\n",mm,Ma,Mi);
//单点修改
//    update(1,4,0);
//    printf("%d\n",Seach(1,4));
    return 0;
}
View Code