线段树 区间更新 区间求和以及最值

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<pre name="code" class="cpp">#include <iostream>#include <string.h>#include <algorithm>#include <stdio.h>#include <math.h>#include <queue>#define MAXN 100010#define inf 0x3f3f3f3fusing namespace std;struct node{    int l,r;//区间[l,r]    int add;//区间的延时标记    int sum;//区间和    int mx; //区间最大值    int mn; //区间最小值}tree[MAXN<<2];//一定要开到4倍多的空间void pushup(int index){    tree[index].sum = tree[index<<1].sum+tree[index<<1|1].sum;    tree[index].mx = max(tree[index<<1].mx,tree[index<<1|1].mx);    tree[index].mn = min(tree[index<<1].mn,tree[index<<1|1].mn);}void pushdown(int index){    //说明该区间之前更新过    //要想更新该区间下面的子区间,就要把上次更新该区间的值向下更新    if(tree[index].add > 0){        //替换原来的值        /*        tree[index<<1].sum = (tree[index<<1].r-tree[index<<1].l+1)*tree[index].add;        tree[index<<1|1].sum = (tree[index<<1|1].r-tree[index<<1|1].l+1)*tree[index].add;        tree[index<<1].mx = tree[index].add;        tree[index<<1|1].mx = tree[index].add;        tree[index<<1].mn = tree[index].add;        tree[index<<1|1].mn = tree[index].add;        tree[index<<1].add = tree[index].add;        tree[index<<1|1].add = tree[index].add;        tree[index].add = 0;*/        //在原来的值的基础上加上val                tree[index<<1].sum += (tree[index<<1].r-tree[index<<1].l+1)*tree[index].add;        tree[index<<1|1].sum +=(tree[index<<1|1].r-tree[index<<1|1].l+1)*tree[index].add;        tree[index<<1].mx += tree[index].add;        tree[index<<1|1].mx += tree[index].add;        tree[index<<1].mn += tree[index].add;        tree[index<<1|1].mn += tree[index].add;        tree[index<<1].add += tree[index].add;        tree[index<<1|1].add += tree[index].add;        tree[index].add = 0;    }}void build(int l,int r,int index){    tree[index].l = l;    tree[index].r = r;    tree[index].add = 0;//刚开始一定要清0    if(l == r){        scanf("%d",&tree[index].sum);        tree[index].mn = tree[index].mx = tree[index].sum;        return ;    }    int mid = (l+r)>>1;    build(l,mid,index<<1);    build(mid+1,r,index<<1|1);    pushup(index);}void updata(int l,int r,int index,int val){    if(l <= tree[index].l && r >= tree[index].r){        /*把原来的值替换成val,因为该区间有tree[index].r-tree[index].l+1        个数,所以区间和 以及 最值为:        */        /*tree[index].sum = (tree[index].r-tree[index].l+1)*val;        tree[index].mn = val;        tree[index].mx = val;        tree[index].add = val;//延时标记*/        //在原来的值的基础上加上val,因为该区间有tree[index].r-tree[index].l+1        //个数,所以区间和 以及 最值为:        tree[index].sum += (tree[index].r-tree[index].l+1)*val;        tree[index].mn += val;        tree[index].mx += val;        tree[index].add += val;//延时标记        return ;    }    pushdown(index);    int mid = (tree[index].l+tree[index].r)>>1;    if(l <= mid){        updata(l,r,index<<1,val);    }    if(r > mid){        updata(l,r,index<<1|1,val);    }    pushup(index);}int query(int l,int r,int index){    if(l <= tree[index].l && r >= tree[index].r){        //return tree[index].sum;        return tree[index].mx;        //return tree[index].mn;    }    pushdown(index);    int mid = (tree[index].l+tree[index].r)>>1;    int ans = 0;    int Max = 0;    int Min = inf;    if(l <= mid){        ans += query(l,r,index<<1);        Max = max(query(l,r,index<<1),Max);        Min = min(query(l,r,index<<1),Min);    }    if(r > mid){        ans += query(l,r,index<<1|1);        Max = max(query(l,r,index<<1|1),Max);        Min = min(query(l,r,index<<1|1),Min);    }    //return ans;    return Max;    //return Min;}int main(){    int n,m,q,x,y,z;    while(~scanf("%d%d",&n,&m)){        build(1,n,1);        while(m--){            scanf("%d",&q);            if(q == 1){                cout<<"查询:(x,y)"<<endl;                scanf("%d %d",&x,&y);                cout<<query(x,y,1)<<endl;            }            else{                cout<<"更新(x,y)为z:"<<endl;                scanf("%d %d %d",&x,&y,&z);                updata(x,y,1,z);                for(int i = 1; i <= n; ++i){                    printf("a[%d] = %d\n",i,query(i,i,1));                }            }        }    }    return 0;}


                                             
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