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  • 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;
    }
    
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  • 原文地址:https://www.cnblogs.com/onionQAQ/p/10507383.html
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