The Preliminary Contest for ICPC China Nanchang National Invitational（2019南昌网络赛）

2019南昌网络赛连接：https://www.jisuanke.com/contest/2290?view=challenges

I. Max answer

题意： 给你一个序列，问任意一个区间，区间和×区间最小值，结果最大为多少。

题解： 区间问题，很容易想到线段树，区间和问题，很容易想到做前缀和相减。所以我们可以用单调栈维护出每个a[i]作为区间最小值所对应的区间 [ l[i],r[i] ]，如果a[i]为正数，则要找出在这个区间的子区间（要包含a[i]）和最大。也就是sum[r]-sum[l-1]最大，也就是sum[r]最大，sum[l-1]最小。所以线段树维护区间内，前缀和的最大值和最小值。对于每个a[i]，如果大于0，找[ l[i]-1,i-1] 的最小值（注意是i-1，这样才能保证a[i]包括在里面），[ i,r[i] ]的最大值。a[i]小于0则反过来。

AC代码：

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <string>
#include <cmath>
#include <ctime>
#include <queue>
#include<stack>
#include <map>
#include <set>
#include<fstream>
#define lowbit(x) x&(-x)
using namespace std;
typedef  long long ll;
typedef  long long LL;
const int N = 5e5 + 5;
const ll INF =1e17;
const int mod = 1000000007;
int a[N], l[N], r[N];
ll sum[N];
struct node
{
	ll mi, ma;
}tree[N<<2];
void up(int i)
{
	tree[i].mi = min(tree[i<<1].mi,tree[i<<1|1].mi);
	tree[i].ma = max(tree[i << 1].ma, tree[i << 1 | 1].ma);
}
void build(int i, int l, int r)
{
	if (l == r) {
		tree[i].mi = tree[i].ma = sum[l];
		return;
	}
	int mid = (l + r) / 2;
	build(i<<1,l,mid);
	build(i<<1|1,mid+1,r);
	up(i);
}
ll query(int i, int L, int R, int l, int r,int flag) // flag 1是最大值 0是最小值
{
	if (l <= L && r >= R) {
		if (flag)
			return tree[i].ma;
		else
			return tree[i].mi;
	}
	int mid = (L+R)/2;
	if (r <= mid) {
		return query(i<<1,L,mid,l,r,flag);
	}
	else if (l > mid) {
		return query(i << 1 | 1, mid + 1, R, l, r, flag);
	}
	else {
		if(flag)
			return max(query(i << 1, L, mid, l, r, flag), query(i << 1 | 1, mid + 1, R, l, r, flag));
		else
			return min(query(i << 1, L, mid, l, r, flag), query(i << 1 | 1, mid + 1, R, l, r, flag));
	}

}
int main()
{
	int n;
	scanf("%d",&n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		sum[i] = sum[i - 1] + a[i];
	}
	a[0] = -INF; a[n + 1] = -INF;
	stack<int>S;
	S.push(0);
	for (int i = 1; i <= n; i++) {//预处理左区间
		while (a[S.top()] >= a[i]) S.pop();
		l[i] = S.top()+1;
		S.push(i);
	}
	while (!S.empty())//注意清空
		S.pop();
	S.push(n+1);
	for (int i = n; i>0; i--) {//右区间
		while (a[S.top()] >= a[i]) S.pop();
		r[i] = S.top()-1;
		S.push(i);
	}
	build(1,0,n);
	ll ans = -1,mi,ma;
	for (int i = 1; i <= n; i++) {//用线段树查询
		if (a[i] >= 0) {
			ma = query(1,0,n,i,r[i],1);
			mi = query(1,0,n,l[i]-1,i-1,0);
			ans = max(ans,a[i]*(ma-mi));
		}
		else {
			mi = query(1, 0, n, i, r[i],0);
			ma = query(1, 0, n, l[i]-1,i-1,1);
			ans = max(ans, a[i] * (mi - ma));
		}
	}
	printf("%lld\n",ans);
	return 0;
}

J. Distance on the tree

题意： 给你一棵树，边上有边权，问两个节点间边权小于K的有多少个。

题解： 树上距离，很容易想到树链剖分，标记好dfs序，每条链的top（链的顶端），每个点的父节点fa，深度dep ，树上两个点的距离就是不断u=fa[top[u]]直到两个点的top一样，具体看代码，不难理解。边权小于k就是主席树了，维护区间dfs序各个边权出现的次数，注意离散化。

AC代码：

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <string>
#include <cmath>
#include <ctime>
#include <queue>
#include<stack>
#include <map>
#include <set>
#include<fstream>
#define lowbit(x) x&(-x)
using namespace std;
typedef  long long ll;
const int N = 2e5 + 5;
const int INF = 0x3f3f3f3f;
int n, m,num;
int head[N],cnt,temp[N],value[N],fa[N],dep[N],siz[N],son[N];
int dfn[N], inv[N],top[N];
int tree[N * 20], ls[N * 20], rs[N * 20], sum[N * 20];
struct node
{
	int u, v,w,next;
}edge[N<<1],q[N];
void add(int u, int v, int w)
{
	cnt++;
	edge[cnt] = { u,v,w,head[u] };
	head[u] = cnt;
}
void dfs1(int u,int f,int w)
{
	fa[u] = f;
	dep[u] = dep[f] + 1;
	siz[u] = 1;
	value[u] = w;
	for (int i = head[u]; ~i; i = edge[i].next) {
		int v = edge[i].v; w = edge[i].w;
		if (v == f)
			continue;
		dfs1(v,u,w);
		siz[u] += siz[v];
		if (siz[v] > siz[son[u]]) {
			son[u] = v;
		}
	}
}
void dfs2(int u, int f)
{
	cnt++;
	dfn[u] = cnt;
	inv[cnt] = u;
	top[u] = f;
	if (son[u] != 0)
		dfs2(son[u],f);
	for (int i = head[u]; ~i; i = edge[i].next) {
		int v = edge[i].v, w = edge[i].w;
		if (v != fa[u]&&v!=son[u])
			dfs2(v,v);
	}
}
void Build(int& o, int l, int r) {
	o = ++cnt;
	sum[o] = 0;
	if (l == r) return;
	int m = (l + r) >> 1;
	Build(ls[o], l, m);
	Build(rs[o], m + 1, r);
}
void update(int& o, int l, int r, int last, int p) {
	o = ++cnt;
	ls[o] = ls[last];
	rs[o] = rs[last];
	if (l == r) {
		sum[o]=sum[last]+1;
		return;
	}
	int m = (l + r) >> 1;
	if (p <= m)  update(ls[o], l, m, ls[last], p);
	if(p>m) update(rs[o], m + 1, r, rs[last], p);
	sum[o] = sum[ls[o]] + sum[rs[o]];
}

int query(int ss, int tt, int l, int r, int k) {
	if (1<=l &&k>=r) 
		return  sum[tt] - sum[ss];
	int m = (l + r) >> 1;
	int ans=0;
	if (1 <= m)
		ans+=query(ls[ss], ls[tt], l, m, k);
	if(k>m)
		ans+=query(rs[ss], rs[tt], m + 1, r, k);
	return ans;
}
int ask(int u, int v, int w)
{
	int fu = top[u], fv = top[v],ans=0;
	while (fu != fv) {
		if (dep[fu] < dep[fv])
			swap(u,v),swap(fu,fv);
		ans += query(tree[dfn[fu]-1], tree[dfn[u]], 1, num+1, w);
		u = fa[fu];
		fu = top[u];
	}
	if (dep[u] < dep[v])
		swap(u, v);
	ans += query(tree[dfn[v]], tree[dfn[u]], 1, num+1, w);
	return ans;
}

int main()
{
	int u, v, w;
	memset(head, -1, sizeof(head));
	cnt = 0;
	scanf("%d%d", &n, &m);
	for (int i = 1; i < n; i++) {
		scanf("%d%d%d", &u, &v, &w);
		temp[i] = w;
		add(u,v,w);
		add(v, u, w);
	}
	dfs1(1,0,INF);
	cnt = 0;
	dfs2(1,1); //树链剖分
	for (int i = 0; i < m; i++) {
		scanf("%d%d%d", &u, &v, &w);
		q[i] = { u,v,w };
		temp[n + i] = w;
	}
	sort(temp+1,temp+n+m);
	 num = unique(temp+1,temp+n+m)-temp-1;
	for (int i = 1; i <= n; i++) {
		value[i] = lower_bound(temp+1,temp+num+1,value[i])-temp; //离散化
	}
	cnt = 0;
	Build(tree[0], 1, num+1);
	for (int i = 1; i <= n; i++) {
		update(tree[i], 1, num+1, tree[i - 1], value[inv[i]]); //建树，按dfs序
	}
	for (int i = 0; i < m; i++) {
		q[i].w = lower_bound(temp + 1, temp + num+1, q[i].w) - temp;
		printf("%d\n",ask(q[i].u,q[i].v,q[i].w));
	}
	return 0;
}