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 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<cstring>
 6 #define MAX 50010
 7 #define inf 100000000
 8 using namespace std;
 9 int a[MAX];
10 struct tree{
11     int l,r,sum,mx,mi,lazy;
12 }f[MAX*4];
13  
14 void pushup(int i){
15     f[i].sum=(f[i*2+1].sum+f[i*2].sum);
16     f[i].mx=max(f[i*2+1].mx,f[i*2].mx);
17     f[i].mi=min(f[i*2+1].mi,f[i*2].mi);
18 }
19  
20 void update(int i,int x)
21 {
22     f[i].sum+=x*(f[i].r-f[i].l+1);
23     f[i].mi+=x;
24     f[i].mx+=x;
25     f[i].lazy+=x;
26     return;
27 }
28  
29 void pushdown(int i){
30     if(f[i].lazy!=0){
31         update(i*2,f[i].lazy);
32         update(i*2+1,f[i].lazy);
33         f[i].lazy=0;
34         return;
35     }
36     else return;
37 }
38  
39 void build(int i,int left,int right){
40     int mid=(left+right)/2;
41     f[i].lazy=0;f[i].mx=0;f[i].mi=inf;
42     f[i].l=left;f[i].r=right;
43     if(left==right){
44         f[i].sum=a[left];
45         f[i].mx=a[left];
46         f[i].mi=a[left];
47         return;
48     }
49     build(i*2,left,mid);
50     build(i*2+1,mid+1,right);
51     pushup(i);
52 }
53  
54 void add(int i,int left,int right,int v){
55     int mid=(f[i].l+f[i].r)/2;
56     if(f[i].l==left&&f[i].r==right){
57         update(i,v);
58         return;
59     }
60     pushdown(i);
61     if(mid>=right)add(i*2,left,right,v);  
62     else if(mid<left)add(i*2+1,left,right,v);
63     else add(i*2,left,mid,v),add(i*2+1,mid+1,right,v);
64     pushup(i);
65 }
66  
67 int query(int i,int left,int right) {
68         int mid=(f[i].l+f[i].r)/2;
69         if(f[i].l==left&&f[i].r==right) return f[i].sum;
70         pushdown(i);
71         if(mid>=right) return query(i*2,left,right);
72         if(mid<left) return query(i*2+1,left,right);
73         return (query(i*2,left,mid)+query(i*2+1,mid+1,right));
74 }
75  
76 int qmax(int i,int left,int right){
77     int mid=(f[i].l+f[i].r)/2;
78     if(f[i].l==left&&f[i].r==right) return f[i].mx;
79     pushdown(i);
80     if(mid>=right) return qmax(i*2,left,right);
81     if(mid<left) return qmax(i*2+1,left,right);
82     return max(qmax(i*2,left,mid),qmax(i*2+1,mid+1,right));
83 }
84 
85 int qmin(int i,int left,int right){
86     int mid=(f[i].l+f[i].r)/2;
87     if(f[i].l==left&&f[i].r==right) return f[i].mi;
88     pushdown(i);
89     if(mid>=right) return qmin(i*2,left,right);
90     if(mid<left) return qmin(i*2+1,left,right);
91     return min(qmin(i*2,left,mid),qmin(i*2+1,mid+1,right));
92 }
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posted @ 2015-09-14 11:06  HTWX  阅读(149)  评论(0)    收藏  举报