LA 3938 动态最大连续和 线段树

题目链接：

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1939

来自：刘汝佳大白书P201.

解题思路：

构造一棵线段树，其中每个结点维护三个值，记录最大前缀和，最大后缀和最大连续和。

最大连续和要么完全在左段，要么完全在右段，要么在跨越中线。就是会是左段的最大后缀和+右段的最大前缀和。。。。

代码也是刘汝佳写的哦

代码：

// LA3938 Ray, Pass me the dishes!
// Rujia Liu
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxn = 500000 + 10;
const int maxnode = 1000000 + 10;
typedef long long LL;
typedef pair<int,int> Interval;

LL prefix_sum[maxn];

LL sum(int L, int R)
{
    return prefix_sum[R] - prefix_sum[L-1];
}

LL sum(Interval p)
{
    return sum(p.first, p.second);
}

Interval better(Interval a, Interval b)
{
    if(sum(a) != sum(b)) return sum(a) > sum(b) ? a : b;
    return a < b ? a : b; // 利用pair自带的字典序
}

int qL, qR;

struct IntervalTree
{
    int max_prefix[maxnode];
    int max_suffix[maxnode];
    Interval max_sub[maxnode];

    void build(int o, int L, int R)
    {
        if(L == R)
        {
            max_prefix[o] = max_suffix[o] = L;
            max_sub[o] = make_pair(L, L);
        }
        else
        {
            int M = L + (R-L)/2;
            // 递归创建子树
            int lc = o*2, rc = o*2+1;
            build(lc, L, M);
            build(rc, M+1, R);

            // 递推max_prefix
            LL v1 = sum(L, max_prefix[lc]);
            LL v2 = sum(L, max_prefix[rc]);
            if(v1 == v2) max_prefix[o] = min(max_prefix[lc], max_prefix[rc]);
            else max_prefix[o] = v1 > v2 ? max_prefix[lc] : max_prefix[rc];

            // 递推max_suffix
            v1 = sum(max_suffix[lc], R);
            v2 = sum(max_suffix[rc], R);
            if(v1 == v2) max_suffix[o] = min(max_suffix[lc], max_suffix[rc]);
            else max_suffix[o] = v1 > v2 ? max_suffix[lc] : max_suffix[rc];

            // 递推max_sub
            max_sub[o] = better(max_sub[lc], max_sub[rc]); // 完全在左子树或者右子树
            max_sub[o] = better(max_sub[o], make_pair(max_suffix[lc], max_prefix[rc])); // 跨越中线
        }
    }

    Interval query_prefix(int o, int L, int R)
    {
        if(max_prefix[o] <= qR) return make_pair(L, max_prefix[o]);
        int M = L + (R-L)/2;
        int lc = o*2, rc = o*2+1;
        if(qR <= M) return query_prefix(lc, L, M);
        Interval i = query_prefix(rc, M+1, R);
        i.first = L;
        return better(i, make_pair(L, max_prefix[lc]));
    }

    Interval query_suffix(int o, int L, int R)
    {
        if(max_suffix[o] >= qL) return make_pair(max_suffix[o], R);
        int M = L + (R-L)/2;
        int lc = o*2, rc = o*2+1;
        if(qL > M) return query_suffix(rc, M+1, R);
        Interval i = query_suffix(lc, L, M);
        i.second = R;
        return better(i, make_pair(max_suffix[rc], R));
    }

    Interval query(int o, int L, int R)
    {
        if(qL <= L && R <= qR) return max_sub[o];
        int M = L + (R-L)/2;
        int lc = o*2, rc = o*2+1;
        if(qR <= M) return query(lc, L, M);
        if(qL > M) return query(rc, M+1, R);
        Interval i1 = query_prefix(rc, M+1, R); // 右半的前缀
        Interval i2 = query_suffix(lc, L, M); // 左半的后缀
        Interval i3 = better(query(lc, L, M), query(rc, M+1, R));
        return better(make_pair(i2.first, i1.second), i3);
    }
};

IntervalTree tree;

int main()
{
    int kase = 0, n, a, Q;
    while(scanf("%d%d", &n, &Q) == 2)
    {
        prefix_sum[0] = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a);
            prefix_sum[i+1] = prefix_sum[i] + a;
        }
        tree.build(1, 1, n);
        printf("Case %d:
", ++kase);
        while(Q--)
        {
            int L, R;
            scanf("%d%d", &L, &R);
            qL = L;
            qR = R;
            Interval ans = tree.query(1, 1, n);
            printf("%d %d
", ans.first, ans.second);
        }
    }
    return 0;
}