ACM——I Hate It（线段树的进化版）

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #define max(a,b) (a > b ? a : b)
 5 #define min(a,b) (a > b ? b : a)
 6 
 7 using namespace std;
 8 
 9 const int MAXN=200010;
10 
11 struct student
12 {
13     int left,right;
14     int nSum;
15 }segTree[MAXN*3];
16 int num[MAXN];
17 
18 
19 
20 
21 void Build(int i,int left,int right)//递归算法，构建一颗线段树
22 {
23     segTree[i].left=left;
24     segTree[i].right=right;
25     if(left==right)
26     {
27         segTree[i].nSum=num[left];
28         return;
29     }
30     int mid=(left+right)>>1;//除2
31     Build(i<<1,left,mid);//i*2 a
32     Build(i<<1|1,mid+1,right);//i*2+1 b
33     segTree[i].nSum=(segTree[i<<1].nSum>segTree[i<<1|1].nSum?segTree[i<<1].nSum:segTree[i<<1|1].nSum);  //取最大值
34 }
35 
36 
37 void U(int i,int tleft,int b)//更新成绩
38 {
39     
40     if(segTree[i].left==tleft&&segTree[i].right==tleft)
41     {
42         segTree[i].nSum=b;
43         return;
44     }
45     int mid=(segTree[i].left+segTree[i].right)>>1;
46 
47     if(tleft<=mid)
48         U(i<<1,tleft,b);
49     else
50         U(i<<1|1,tleft,b);
51 
52     segTree[i].nSum=(segTree[i].nSum>b?segTree[i].nSum:b);
53 
54 }
55 
56 
57 int Q(int i,int left,int right)
58 {
59     if(segTree[i].left==left&&segTree[i].right==right)
60     {
61         return segTree[i].nSum;
62     }
63     int middle=(segTree[i].left+segTree[i].right)>>1;
64     if(left>middle)
65     {
66         return Q(i<<1|1,left,right);
67     }
68     else if(right<=middle)
69     {
70         return Q(i<<1,left,right);
71     }
72     else
73     {
74         int a=Q(i<<1,left,middle);
75         int b=Q(i<<1|1,middle+1,right);
76         return a>b?a:b;
77     }
78 }
79 
80 
81 int main()
82 {
83     int n,t,x,y;
84     char C;
85     while(scanf("%d%d",&n,&t)!=EOF){
86         for(int i=1; i<= n; i++)scanf("%d",&num[i]);
87         Build(1,1,n);
88         while(t--){
89             getchar();   //吸收掉多余的换行键
90             scanf("%c%d%d",&C,&x,&y);
91 
92             if(C == ‘U‘)
93                 U(1,x,y);
94             else 
95                 printf("%d\n",Q(1,x,y));    
96         }    
97     }    
98     return 0;
99 }