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Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 51763 Accepted Submission(s): 20297
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #define max(a, b) a>b?a:b 5 using namespace std; 6 const int MAXN = 200020; 7 struct Node 8 { 9 int max, left, right; 10 } tree[MAXN * 3]; 11 int n, m, num[MAXN]; 12 int Build(int root, int left, int right) 13 { 14 int mid; 15 tree[root].left = left; 16 tree[root].right = right; 17 if(left == right) // 叶子; 18 return tree[root].max = num[left]; 19 mid = (left + right) / 2; 20 int a, b; 21 a = Build(2*root, left, mid); 22 b = Build(2*root + 1, mid + 1, right); 23 return tree[root].max = max(a, b); 24 } 25 int Update(int root, int pos, int val) 26 { 27 if(pos < tree[root].left || tree[root].right < pos) //该点不在子段中 28 return tree[root].max; 29 if(pos == tree[root].left && pos == tree[root].right) //叶子; 30 return tree[root].max = val; 31 int a, b; 32 a = Update(2*root, pos, val); 33 b = Update(2*root+1, pos, val); 34 tree[root].max = max(a, b); 35 return tree[root].max; 36 } 37 int Find(int root, int left, int right) 38 { 39 int mid; 40 if(tree[root].left > right || tree[root].right < left) //没有相交线段; 41 return 0; 42 if(tree[root].left >= left && tree[root].right <= right) // 有一段连续的线段包含在查找区间内; 43 return tree[root].max; 44 int a, b; 45 a = Find(root*2, left, right); 46 b = Find(root*2 + 1, left, right); 47 return max(a, b); 48 } 49 int main() 50 { 51 while(~scanf("%d %d", &n, &m)) 52 { 53 for(int i = 1; i <= n; i++) 54 scanf("%d", &num[i]); 55 Build(1, 1, n); 56 char str[10]; int a, b; 57 for(int i = 1; i <= m; i++) 58 { 59 cin >> str >> a >> b; 60 if(str[0] == ‘Q‘) 61 printf("%d\n", Find(1, a, b)); 62 else 63 { 64 num[a] = b; 65 Update(1, a, b); 66 } 67 } 68 } 69 return 0; 70 }
1 http://blog.csdn.net/metalseed/article/details/8039326#
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原文地址:http://www.cnblogs.com/fengshun/p/4746999.html