hdu1754 I Hate It

题目链接：hdu1754 I Hate It

思路

单点更新。区间最大。

小结

又WA了一发。设查询区间为[QL, QR],当前节点为rt，对应区间为[L, R], mid=(L+R)/2。按照我的写法，当QL小于等于mid时，递归左区间，当QR大于mid时，递归右区间。这里应该是两个if，而我写成了if-else，含义就错了。

代码

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>using namespace std;const int maxn = 200000+5;int tree[maxn<<2];void build(int rt, int l, int r){    if(l==r){        scanf("%d", &tree[rt]);    }    else{        int mid = (l+r)>>1;        build(rt<<1, l, mid);        build(rt<<1|1, mid+1, r);        tree[rt] = max(tree[rt<<1], tree[rt<<1|1]);    }}void Set(int rt, int id, int l, int r, int x){    if(l==r && l==id){        tree[rt] = x;    }    else{        int mid = (l+r)>>1;        if(id<=mid)             Set(rt<<1, id, l, mid, x);        else             Set(rt<<1|1, id, mid+1, r, x);        tree[rt] = max(tree[rt<<1], tree[rt<<1|1]);    }}int query(int rt, int a, int b, int l, int r){    if(a<=l && r<=b){       return tree[rt];     }    else{        int mid = (l+r)>>1;        int ans = -1;        if(a<=mid)             ans =  max(ans, query(rt<<1, a, b, l, mid));        if(mid<b)            ans = max(ans, query(rt<<1|1, a, b, mid+1, r));        return ans;    }}int main(){    int n, m, a, b;    char opt[10];    while(scanf("%d%d", &n, &m)!=EOF){        build(1, 1, n);        for(int i=0; i<m; ++i){            scanf("%s%d%d", opt, &a, &b);            if(opt[0]=='U'){                Set(1, a, 1, n, b);            }            if(opt[0]=='Q'){                printf("%d\n", query(1, a, b, 1, n));            }        }    }    return 0;}