SPOJ GSS3 Can you answer these queries III

题意：和上一篇中的意思一样，只是多了一种操作：单点更新 ，查找到叶子位置，改变一下就行~~~

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=19937

代码如下

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <algorithm>using namespace std;#define maxn  50010#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1struct SegNode{    int lsum,rsum,sum,ans;};SegNode tree[maxn<<2];void PushUp(int rt){    tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;    tree[rt].lsum = max(tree[rt<<1].lsum , tree[rt<<1].sum + tree[rt<<1|1].lsum);    tree[rt].rsum = max(tree[rt<<1|1].rsum , tree[rt<<1|1].sum + tree[rt<<1].rsum);    tree[rt].ans = max(max(tree[rt<<1].ans , tree[rt<<1|1].ans),tree[rt<<1].rsum + tree[rt<<1|1].lsum);}void build(int l,int r,int rt){    if(l == r){        scanf("%d",&tree[rt].sum);        tree[rt].lsum = tree[rt].rsum = tree[rt].ans = tree[rt].sum;        return ;    }    int m = (l + r)/2;    build(lson);    build(rson);    PushUp(rt);}void update(int pos,int val,int l,int r,int rt){    if(l == r){        tree[rt].lsum = tree[rt].rsum = tree[rt].ans = tree[rt].sum = val;        return ;    }    int m = (l + r)/2;    if(pos <= m) update(pos , val , lson);    else update(pos , val , rson);    PushUp(rt);}SegNode query(int L,int R,int l,int r,int rt){    if(L <= l && r <= R){        return tree[rt];    }    int m = (l + r)/2;    if(R <= m) return query(L , R , lson);    else if(L > m) return query(L , R , rson);    else{        SegNode result,left,right;        left = query(L , m ,lson);        right = query(m + 1, R, rson);        result.sum = left.sum + right.sum;        result.lsum = max(left.lsum , left.sum + right.lsum);        result.rsum = max(right.rsum , right.sum + left.rsum);        result.ans = max(max(left.ans , right.ans), left.rsum + right.lsum);        return result;    }}int n,m,op,a,b;int main(){    while(~scanf("%d",&n)){        build(1 , n , 1);        scanf("%d",&m);        while(m --){            scanf("%d %d %d",&op,&a,&b);            if(op == 1)                printf("%d\n",query(a , b, 1 , n , 1).ans);            else                update(a, b, 1, n, 1);        }    }    return 0;}