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Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 64834 Accepted Submission(s): 25168
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <stack> #include <queue> #define maxn 1000007 #define INF 0x3f3f3f3f typedef long long LL; using namespace std; #define lson rt<<1 #define rson rt<<1|1 int A[maxn]; struct node { int l, r, Max; } tree[maxn<<2];///4倍 void build(int l ,int r, int rt)///建立一棵树 { tree[rt].l=l; tree[rt].r=r; if(l==r) { tree[rt].Max=A[l]; return; } int mid=(l+r)>>1; build(l, mid, lson); build(mid+1, r, rson); tree[rt].Max=max(tree[lson].Max,tree[rson].Max); } ///更新函数 void update(int pos, int val, int rt) { tree[rt].Max=max(tree[rt].Max,val); if(tree[rt].l==tree[rt].r) { tree[rt].Max=val; return; } int mid=(tree[rt].l+tree[rt].r)>>1; if(pos<=mid)update(pos, val, lson); else update(pos, val, rson); } ///查询函数 int query(int l, int r, int rt) { if(tree[rt].l==l&&tree[rt].r==r) return tree[rt].Max; int mid=(tree[rt].l+tree[rt].r)>>1; if(r<=mid)return query(l, r, lson);///求在左儿子那一边的最大值 else if(l>mid)return query(l, r, rson);///求在右儿子那一边的最大值 else { int ans1, ans2;///求左右两边的最大值,再返回最大的 ans1=query(l, mid, lson); ans2=query(mid+1, r, rson); return max(ans1, ans2); } } int main() { int n, q; while(~scanf("%d%d", &n, &q)) { for(int i=1; i<=n; i++) scanf("%d", &A[i]); build(1, n, 1); int a, b; char op[20]; while(q--) { scanf("%s %d %d", op, &a, &b); if(op[0]==‘Q‘) { printf("%d\n", query(a, b, 1)); continue; } update(a, b, 1); } } return 0; }
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原文地址:http://www.cnblogs.com/w-y-1/p/5738424.html