hdu 5306 Gorgeous Sequence

         对线段树进行三种操作：

          区间求和。


          区间求最大值。


          将节点值更新为当前值与给定值中的最小值。

怎么做？代码中有详细注解，看看就懂了。

#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<iostream>#include<algorithm>#include<bitset>#include<climits>#include<list>#include<iomanip>#include<stack>#include<set>using namespace std;typedef long long ll;/************Read**********/  char *ch, *ch1, buf[40*1024000+5], buf1[40*1024000+5];    void read(int &x)  {      for (++ch; *ch <= 32; ++ch);      for (x = 0; '0' <= *ch; ch++)    x = x * 10 + *ch - '0';  }  /**************************/  struct Tree{int l,r,num,mx,flag;//这个结点当前管控num个数，这是用来快速求和用的//mx:这个结点的最大值//flag:这个结点的修改值 ll sum;//sum:这个结点的和 }tree[1000000<<2];void makeflag(int k,int flag){if(tree[k].flag!=0&&tree[k].flag<=flag)return;tree[k].flag=flag;tree[k].sum+=ll(tree[k].r-tree[k].l+1-tree[k].num)*flag;//num发挥求和的作用，若管控不了这个结点的所有数，则会加上修改值 if(tree[k].num<tree[k].r-tree[k].l+1)tree[k].mx=flag;tree[k].num=tree[k].r-tree[k].l+1;//修改完毕后，已经管控了这个结点所有数 }void pushdown(int k)//更新k的儿子 {if(tree[k].flag==0)return;makeflag(k<<1,tree[k].flag);makeflag(k<<1|1,tree[k].flag);}void pushup(int k)//用儿子更新k {tree[k].mx=max(tree[k<<1].mx,tree[k<<1|1].mx);tree[k].num=tree[k<<1].num+tree[k<<1|1].num;tree[k].sum=tree[k<<1].sum+tree[k<<1|1].sum;}void build(int l,int r,int k){tree[k].l=l;tree[k].r=r;tree[k].flag=0;if(l==r){read(tree[k].flag);tree[k].num=1;tree[k].mx=tree[k].sum=tree[k].flag;return;}int m=l+r>>1;build(l,m,k<<1);build(m+1,r,k<<1|1);pushup(k);}void dfs(int k,int flag){if(tree[k].mx<=flag)return;if(tree[k].flag>=flag)tree[k].flag=0;if(tree[k].l==tree[k].r){tree[k].mx=tree[k].sum=tree[k].flag;tree[k].num=tree[k].flag!=0;return;}pushdown(k);dfs(k<<1,flag);dfs(k<<1|1,flag);pushup(k);}void update(int l,int r,int flag,int k){if(tree[k].mx<=flag)return;if(l==tree[k].l&&r==tree[k].r){dfs(k,flag);//更新这个结点的后代 makeflag(k,flag);//更新这个结点 return;}int m=tree[k].l+tree[k].r>>1;pushdown(k);if(r<=m)update(l,r,flag,k<<1);else if(l>m)update(l,r,flag,k<<1|1);else{update(l,m,flag,k<<1);update(m+1,r,flag,k<<1|1);}pushup(k);}Tree seek(int l,int r,int k){if(l==tree[k].l&&r==tree[k].r)return tree[k];int m=tree[k].l+tree[k].r>>1;pushdown(k);if(r<=m)return seek(l,r,k<<1);if(l>m)return seek(l,r,k<<1|1);Tree t1=seek(l,m,k<<1);Tree t2=seek(m+1,r,k<<1|1);t1.mx=max(t1.mx,t2.mx);t1.sum+=t2.sum;return t1;}int main(){ch = buf - 1;      ch1 = buf1 - 1;      fread(buf, 1, 1000 * 35 * 1024, stdin);  int T;read(T);while(T--){int n,m;read(n);read(m);build(1,n,1);while(m--){int x,l,r;read(x);read(l);read(r);if(x==0){read(x);update(l,r,x,1);}else{if(x==1)printf("%d\n",seek(l,r,1).mx);elseprintf("%I64d\n",seek(l,r,1).sum);}}}}

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 521    Accepted Submission(s): 110

Problem Description

There is a sequence a of length n. We use ai to denote the i-th element in this sequence. You should do the following three types of operations to this sequence.

0 x y t: For every x≤i≤y, we use min(ai,t) to replace the original ai's value.
1 x y: Print the maximum value of ai that x≤i≤y.
2 x y: Print the sum of ai that x≤i≤y.

 

Input

The first line of the input is a single integer T, indicating the number of testcases. 

The first line contains two integers n and m denoting the length of the sequence and the number of operations.

The second line contains n separated integers a1,…,an (∀1≤i≤n,0≤ai<231).

Each of the following m lines represents one operation (1≤x≤y≤n,0≤t<231).

It is guaranteed that T=100, ∑n≤1000000, ∑m≤1000000.

 

Output

For every operation of type 1 or 2, print one line containing the answer to the corresponding query.

 

Sample Input

15 51 2 3 4 51 1 52 1 50 3 5 31 1 52 1 5

 

Sample Output

515312
Hint
Please use efficient IO method
 

 

Source

2015 Multi-University Training Contest 2