题意:
求树上最长上升路径
解析:
树状数组版: 998ms
edge[u][w] 代表以u为一条路的终点的小于w的最长路径的路的条数
· 那么edge[v][w] = max(edge[u][w-1]) + 1;
因为w最小是0 所以所有的w都+1
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6+, INF = 0x7fffffff;
int n, m, maxx = -INF;
map<int, int> edge[maxn];
int lowbit(int x)
{
return x & -x;
}
int qp(int u, int w)
{
int ret = ;
for(int i=w; i>; i-=lowbit(i))
ret = max(ret, edge[u][i]);
return ret;
} int build(int u, int w, int ans)
{
while(w)
{
edge[u][w] = max(edge[u][w], ans);
w += lowbit(w);
} } int main()
{
int u, v, w;
cin >> n >> m;
for(int i=; i<m; i++)
{
cin >> u >> v >> w;
maxx = max(maxx, w);
build(v, +, qp(u, w)+);
}
int max_ret = -INF;
for(int i=; i<=n; i++)
max_ret = max(max_ret, qp(i, ));
cout << max_ret << endl; return ;
}
主席树: 108ms
每棵树都建立100000个结点 每次更新小于w的结点的sum
#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define pd(a) printf("%d\n", a);
#define plld(a) printf("%lld\n", a);
#define pc(a) printf("%c\n", a);
#define ps(a) printf("%s\n", a);
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _ ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 1e5 + , INF = 0x7fffffff;
int n, m, cnt, root[maxn], a[maxn], x, y, k;
struct node{int l, r, sum;}T[maxn*];
void update(int l, int r, int& x, int w, int ci)
{
if(!x) x = ++cnt; T[x].sum = max(T[x].sum, ci);
if(l == r) return;
int mid = (l + r) / ;
if(mid >= w) return update(l, mid, T[x].l, w, ci);
else return update(mid+, r, T[x].r, w, ci);
} int query(int l, int r, int x, int k)
{
if(l == r) return T[x].sum;
int mid = (l + r)/;
if(mid >= k) return query(l, mid, T[x].l, k);
else return max(T[T[x].l].sum, query(mid+, r, T[x].r, k));
} int main()
{
int u, v, w, ret = -INF;
rd(n), rd(m);
rep(i, , m)
{
rd(u), rd(v), rd(w);
w++;
int tmp = query(, , root[u], w-) + ;
update(, , root[v], w, tmp);
ret = max(ret, tmp);
}
cout<< ret <<endl; return ;
}