Pathwalks CodeForces

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+10, 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 = 0;
    for(int i=w; i>0; 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=0; i<m; i++)
    {
        cin >> u >> v >> w;
        maxx = max(maxx, w);
        build(v, +1, qp(u, w)+1);
    }
    int max_ret = -INF;
    for(int i=1; i<=n; i++)
        max_ret = max(max_ret, qp(i, 100001));
    cout << max_ret << endl;


    return 0;
}

 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
", a);
#define plld(a) printf("%lld
", a);
#define pc(a) printf("%c
", a);
#define ps(a) printf("%s
", 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 + 6, INF = 0x7fffffff;
int n, m, cnt, root[maxn], a[maxn], x, y, k;
struct node{int l, r, sum;}T[maxn*40];
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) / 2;
    if(mid >= w) return update(l, mid, T[x].l, w, ci);
    else return update(mid+1, 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)/2;
    if(mid >= k) return query(l, mid, T[x].l, k);
    else return max(T[T[x].l].sum, query(mid+1, r, T[x].r, k));
}

int main()
{
    int u, v, w, ret = -INF;
    rd(n), rd(m);
    rep(i, 0, m)
    {
        rd(u), rd(v), rd(w);
        w++;
        int tmp = query(1, 100001, root[u], w-1) + 1;
        update(1, 100001, root[v], w, tmp);
        ret = max(ret, tmp);
    }
    cout<< ret <<endl;


    return 0;
}