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  • 线段树(区间合并) LA 3989 "Ray, Pass me the dishes!"

    题目传送门

    题意:动态最大连续子序列和,静态的题目

    分析:nlogn的归并思想。线段树维护结点的三个信息,最大前缀和,最大后缀和,该区间的最大和的两个端点,然后答案是三个的better。书上用pair保存端点,用自带的<来得到最优。

    #include <bits/stdc++.h>
    using namespace std;
    
    #define lson l, mid, rt << 1
    #define rson mid + 1, r, rt << 1 | 1
    typedef long long ll;
    typedef pair<int, int> P;
    const int N = 5e5 + 5;
    ll sum[N];
    struct	ST	{
    	int pre[N<<2], suf[N<<2];
    	P sub[N<<2];
    	ll get_sum(P p)	{
    		return sum[p.second] - sum[p.first-1];
    	}
    	P better(P a, P b)	{
    		ll v1 = get_sum (a), v2 = get_sum (b);
    		if (v1 != v2)	return v1 > v2 ? a : b;
    		else	return a < b ? a : b;
    	}
    	void push_up(int l, int r, int rt)	{
    		pre[rt] = better (make_pair (l, pre[rt<<1]), make_pair (l, pre[rt<<1|1])).second;	//该区间的最大前缀
    		suf[rt] = better (make_pair (suf[rt<<1], r), make_pair (suf[rt<<1|1], r)).first;	//该区间的最大后缀
    		sub[rt] = better (sub[rt<<1], sub[rt<<1|1]);		//该区间的最大连续和:max (左前缀,右后缀+左前缀,右后缀)
    		sub[rt] = better (sub[rt], make_pair (suf[rt<<1], pre[rt<<1|1]));	//不一定就是最大前缀或最大后缀
    	}
    	void build(int l, int r, int rt)	{
    		if (l == r)	{
    			pre[rt] = suf[rt] = l;
    			sub[rt] = make_pair (l, l);
    			return ;
    		}
    		int mid = (l + r) >> 1;
    		build (lson);	build (rson);
    		push_up (l, r, rt);
    	}
    	P query_pre(int ql, int qr, int l, int r, int rt)	{
    		if (pre[rt] <= qr)	return make_pair (l, pre[rt]);
    		int mid = (l + r) >> 1;
    		if (qr <= mid)	return query_pre (ql, qr, lson);
    		P p = query_pre (ql, qr, rson);	p.first = l;
    		return better (p, make_pair (l, pre[rt<<1]));
    	}
    	P query_suf(int ql, int qr, int l, int r, int rt)	{
    		if (suf[rt] >= ql)	return make_pair (suf[rt], r);
    		int mid = (l + r) >> 1;
    		if (ql > mid)	return query_suf (ql, qr, rson);
    		P p = query_suf (ql, qr, lson);	p.second = r;
    		return better (p, make_pair (suf[rt<<1|1], r));
    	}
    	P query(int ql, int qr, int l, int r, int rt)	{
    		if (ql <= l && r <= qr)	return sub[rt];
    		int mid = (l + r) >> 1;
    		if (qr <= mid)	return query (ql, qr, lson);
    		if (ql > mid)	return query (ql, qr, rson);
    		P p1 = query_suf (ql, qr, lson);				//ql <= mid < qr
    		P p2 = query_pre (ql, qr, rson);				//和push_up一样
    		P p3 = better (query (ql, qr, lson), query (ql, qr, rson));	//该区间的最大连续和:max (左前缀,右后缀+左前缀,右后缀)
    		return better (p3, make_pair (p1.first, p2.second));
    	}
    }st;
    
    int main(void)	{
    	int n, q, cas = 0;
    	while (scanf ("%d%d", &n, &q) == 2)	{
    		for (int i=1; i<=n; ++i)	{
    			scanf ("%lld", &sum[i]);	sum[i] += sum[i-1];
    		}
    		st.build (1, n, 1);
    		printf ("Case %d:
    ", ++cas);
    		int ql, qr;
    		while (q--)	{
    			scanf ("%d%d", &ql, &qr);
    			P ans = st.query (ql, qr, 1, n, 1);
    			printf ("%d %d
    ", ans.first, ans.second);
    		}
    	}
    
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Running-Time/p/5033694.html
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