• Divide by Zero 2018 and Codeforces Round #474 (Div. 1 + Div. 2, combined)


    思路:把边看成点,然后每条边只能从下面的边转移过来,我们将边按照u为第一关键字,w为第二关键字排序,这样就能用线段树维护啦。

      1 #include<bits/stdc++.h>
      2 #define LL long long
      3 #define fi first
      4 #define se second
      5 #define mk make_pair
      6 #define pii pair<int,int>
      7 using namespace std;
      8 
      9 const int N = 1e6 + 7;
     10 const int M = 100 + 7;
     11 const int inf = 0x3f3f3f3f;
     12 const LL INF = 0x3f3f3f3f3f3f3f3f;
     13 const int mod = 1e9 + 7;
     14 
     15 
     16 int n, m, hs[N];
     17 
     18 struct node {
     19     int u, v, w;
     20     node(int u = 0, int v = 0, int w = 0) {
     21         this -> u = u;
     22         this -> v = v;
     23         this -> w = w;
     24     }
     25     bool operator < (const node &rhs) const {
     26         if(u == rhs.u) {
     27             return w < rhs.w;
     28         } else {
     29             return u < rhs.u;
     30         }
     31     }
     32 } edge[N], edge2[N];
     33 
     34 
     35 struct seg_tree {
     36     struct node {
     37         int mx, l, r;
     38     }a[N << 2];
     39 
     40     void build(int l, int r, int rt) {
     41         a[rt].l = l, a[rt].r = r;
     42         if(l == r) return;
     43         int mid = (l + r) >> 1;
     44         build(l, mid, rt << 1);
     45         build(mid + 1, r, rt << 1 | 1);
     46     }
     47 
     48     void update(int pos, int rt, int v) {
     49         int l = a[rt].l, r = a[rt].r;
     50         if(l == r) {
     51             a[rt].mx = max(a[rt].mx, v);
     52             return;
     53         }
     54 
     55         int mid = (l + r) >> 1;
     56 
     57         if(pos <= mid)
     58             update(pos, rt << 1, v);
     59         else
     60             update(pos, rt << 1 | 1, v);
     61 
     62         a[rt].mx = max(a[rt << 1].mx, a[rt << 1 | 1].mx);
     63     }
     64 
     65     int query(int L, int R, int rt) {
     66         int l = a[rt].l, r = a[rt].r;
     67         if(l >= L && r <= R) {
     68             return a[rt].mx;
     69         }
     70 
     71         int mid = (l + r) >> 1;
     72         int ans = 0;
     73 
     74         if(L <= mid)
     75             ans = max(ans, query(L, R, rt << 1));
     76         if(R > mid)
     77             ans = max(ans, query(L, R, rt << 1 | 1));
     78 
     79         return ans;
     80     }
     81 }seg;
     82 
     83 int main() {
     84 
     85     scanf("%d%d", &n, &m);
     86 
     87     for(int i = 1; i <= m; i++) {
     88         scanf("%d", &edge[i].u);
     89         scanf("%d", &edge[i].v);
     90         scanf("%d", &edge[i].w);
     91         edge2[i].u = edge[i].u;
     92         edge2[i].v = edge[i].v;
     93         edge2[i].w = edge[i].w;
     94     }
     95 
     96     sort(edge2 + 1, edge2 + m + 1);
     97     int ans = 0;
     98 
     99     seg.build(1, m, 1);
    100     for(int i = m; i >= 1; i--) {
    101         int l = lower_bound(edge2 + 1, edge2 + m + 1, node(edge[i].v, 0, edge[i].w + 1)) - edge2;
    102         int r = lower_bound(edge2 + 1, edge2 + m + 1, node(edge[i].v + 1, 0, -1)) - edge2 - 1;
    103         int ret = 1;
    104         if(l <= r) {
    105             ret = max(ret, 1 + seg.query(l, r, 1));
    106         }
    107         int pos = lower_bound(edge2 + 1, edge2 + m + 1, edge[i]) - edge2;
    108         seg.update(pos, 1, ret);
    109     }
    110 
    111     printf("%d
    ", seg.a[1].mx);
    112     return 0;
    113 }
    114 
    115 /*
    116 */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/8761154.html
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