洛谷P4513 小白逛公园

区间最大子段和模板题。。

维护四个数组:prefix, suffix, sum, tree

假设当前访问节点为cur

  • prefix[cur]=max(prefix[lson],sum[lson]+preifx[rson])
  • suffix[cur]=max(suffix[rson],sum[rson]+suffix[lson])
  • sum[cur]=sum[lson]+sum[rson]
  • tree[cur]=max(tree[lson], tree[rson], suffix[lson]+suffix[rson])

在query的时候要注意合并,其实和pushup差不多,需要返回一个节点

// luogu-judger-enable-o2
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C yql){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % yql)if(p & 1)ans = 1LL * x * ans % yql;
    return ans;
}
const int N = 500005;
struct Node{
    int prefix, suffix, tree, sum;
    Node(int p, int s, int t, int u): prefix(p), suffix(s), tree(t), sum(u){}
};
int n, m, a[N];
int prefix[N<<2], suffix[N<<2], sum[N<<2], tree[N<<2];

void push_up(int treeIndex){
    int lson = treeIndex << 1, rson = treeIndex << 1 | 1;
    prefix[treeIndex] = max(prefix[lson], sum[lson] + prefix[rson]);
    suffix[treeIndex] = max(suffix[rson], sum[rson] + suffix[lson]);
    sum[treeIndex] = sum[lson] + sum[rson];
    tree[treeIndex] = max(tree[lson], tree[rson], suffix[lson] + prefix[rson]);
}

void buildTree(int treeIndex, int l, int r){
    if(l == r){
        tree[treeIndex] = prefix[treeIndex] = suffix[treeIndex] = sum[treeIndex] = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    buildTree(treeIndex << 1, l, mid);
    buildTree(treeIndex << 1 | 1, mid + 1, r);
    push_up(treeIndex);
}

void modify(int treeIndex, int l, int r, int k, int e){
    if(l == r){
        tree[treeIndex] = prefix[treeIndex] = suffix[treeIndex] = sum[treeIndex] = e;
        return;
    }
    int mid = (l + r) >> 1;
    if(k <= mid) modify(treeIndex << 1, l, mid, k, e);
    else modify(treeIndex << 1 | 1, mid + 1, r, k, e);
    push_up(treeIndex);
}

Node query(int treeIndex, int l, int r, int queryL, int queryR){
    if(l == queryL && r == queryR){
        return Node(prefix[treeIndex], suffix[treeIndex], tree[treeIndex], sum[treeIndex]);
    }
    int mid = (l + r) >> 1;
    if(queryL > mid) return query(treeIndex << 1 | 1, mid + 1, r, queryL, queryR);
    else if(queryR <= mid) return query(treeIndex << 1, l, mid, queryL, queryR);
    Node lr = query(treeIndex << 1, l, mid, queryL, mid);
    Node rr = query(treeIndex << 1 | 1, mid + 1, r, mid + 1, queryR);
    Node ret = Node(max(lr.prefix, lr.sum + rr.prefix),
            max(rr.suffix, rr.sum + lr.suffix),
            max(lr.tree, rr.tree, lr.suffix + rr.prefix),
            lr.sum + rr.sum);
    return ret;
}

int main(){

    n = read(), m = read();
    for(int i = 1; i <= n; i ++) a[i] = read();
    buildTree(1, 1, n);
    while(m --){
        int opt = read(), p = read(), q = read();
        if(opt == 1){
            if(p > q) swap(p, q);
            printf("%d
", query(1, 1, n, p, q).tree);
        }
        else if(opt == 2){
            modify(1, 1, n, p, q);
        }
    }
    return 0;
}
原文地址:https://www.cnblogs.com/onionQAQ/p/10507383.html