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http://acm.hdu.edu.cn/showproblem.php?pid=1754
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 49750 Accepted Submission(s): 19544
#include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #define N 200010 #define Lson root<<1, L, tree[root].Mid() #define Rson root<<1|1, tree[root].Mid() + 1, R using namespace std; struct Tree { int L, R, Max; int Mid() { return (L + R) / 2; } }tree[N * 4]; int al[N]; void Build(int root, int L, int R) { tree[root].L = L, tree[root].R = R; if(L == R) { tree[root].Max = al[L]; return ; } Build(Lson); Build(Rson); tree[root].Max = max(tree[root<<1].Max, tree[root<<1|1].Max); } void Update(int root, int k, int e) { if(tree[root].L == tree[root].R) { tree[root].Max = e; return ; } if(k <= tree[root].Mid()) Update(root<<1, k, e); else Update(root<<1|1, k, e); tree[root].Max = max(tree[root<<1].Max, tree[root<<1|1].Max);//更新k点后,还要更新节点的最大值,上一区间节点的最大值取下面左右两个区间节点的最大值 } int Query(int root, int L, int R) { if(tree[root].L == L && tree[root].R == R) return tree[root].Max; if(R <= tree[root].Mid()) return Query(root<<1, L, R); else if(L > tree[root].Mid()) return Query(root<<1|1, L, R); else return max(Query(Lson), Query(Rson)); } int main() { int n, m, a, b, i; char s[10]; while(~scanf("%d%d", &n, &m)) { for(i = 1 ; i <= n ; i++) scanf("%d", &al[i]); Build(1, 1, n); while(m--) { scanf("%s%d%d", s, &a, &b); if(s[0] == ‘U‘) Update(1, a, b); else printf("%d\n", Query(1, a, b)); } } return 0; } /* 10 2 1 2 3 4 5 10 6 7 8 9 U 6 8 Q 1 6 */
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原文地址:http://www.cnblogs.com/qq2424260747/p/4701049.html