hdu1754 I Hate It

来源：互联网发布：交互式课件制作软件编辑：程序博客网时间：2024/06/12 19:00

题意：n个数，m个操作。操作可以是更新某个数的值或查询某个区间的最大值。

思路：线段树（单点更新）。节点维护一个区间最值的信息。

#include <iostream>    #include <stdio.h>    #include <cmath>    #include <algorithm>    #include <iomanip>    #include <cstdlib>    #include <string>    #include <memory.h>    #include <vector>    #include <queue>    #include <stack>    #include <map>  #include <set>  #include <ctype.h>    using namespace std;  int num[200010];struct node{int l;int r;int v;};node tree[800010];int build_tree(int n,int l,int r){tree[n].l=l;tree[n].r=r;if(l!=r){int mid=(l+r)/2;int v1=build_tree(n*2,l,mid);int v2=build_tree(n*2+1,mid+1,r);tree[n].v= max(v1,v2);}else{tree[n].v=num[l];}return tree[n].v;}void update(int pos,int n,int v){int mid=(tree[n].l+tree[n].r)/2;if(tree[n].l==tree[n].r){tree[n].v=v;return;}if(mid>=pos){update(pos,n*2,v);}else{update(pos,n*2+1,v);}tree[n].v=max(tree[2*n].v,tree[2*n+1].v);}int query(int n,int l,int r){if(tree[n].l==l&&tree[n].r==r){return tree[n].v;}int mid=(tree[n].l+tree[n].r)/2;if(mid>=r){return query(n*2,l,r);}else{if(mid<l){return query(n*2+1,l,r);}else{return max(query(n*2,l,mid),query(n*2+1,mid+1,r));}}}int main(){int n,m;while(cin>>n>>m){for(int i=1;i<=n;i++){scanf("%d",&num[i]);}build_tree(1,1,n);for(int i=1;i<=m;i++){char c;int a,b;scanf("%c%c %d %d",&c,&c,&a,&b);if(c=='U'){update(a,1,b);}else{int ans=query(1,a,b);printf("%d\n",ans);}}}return 0;}

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