• Jam's problem again HDU


    Jam's problem again

    HDU - 5618

     一维快排、二维CDQ、三维树状数组

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 const int maxn = 100010;
     4 
     5 struct Node{
     6     int x, y, z, id;
     7     Node(int x = 0, int y = 0, int z = 0, int id = 0) : 
     8         x(x), y(y), z(z), id(id) {}
     9     bool operator < (const Node &a)const{
    10         if(x != a.x) return x < a.x;
    11         if(y != a.y) return y < a.y;
    12         return z < a.z;
    13     }
    14     bool operator == (const Node &a)const{
    15         return x == a.x && y == a.y && z == a.z;
    16     }
    17 }q[maxn], tq[maxn];
    18 int ans[maxn];
    19      
    20 bool cmp(Node &a, Node &b){
    21     return a.id < b.id;
    22 }
    23 
    24 struct BIT{
    25     int n, c[maxn];
    26     void init(int _n){
    27         n = _n;
    28         memset(c, 0, sizeof c);
    29     }
    30     void add(int i, int v){
    31         for(; i <= n; i += i & -i) c[i] += v;
    32     }
    33     int sum(int i){
    34         int res = 0;
    35         for(; i; i -= i & -i) res += c[i];
    36         return res;
    37     }
    38     
    39 }bit;
    40 void CDQ(int l, int r){
    41     if(l == r) return;
    42     int m = l + r >> 1;
    43     CDQ(l, m); CDQ(m + 1, r);
    44     int f1 = l, f2 = m + 1;
    45     int i = l;
    46     while(f1 <= m || f2 <= r){
    47         if(f2 > r || (f1 <= m && q[f1].y <= q[f2].y)){
    48             bit.add(q[f1].z, 1);
    49             tq[i++] = q[f1++];
    50         }else{
    51             ans[q[f2].id] += bit.sum(q[f2].z);
    52             tq[i++] = q[f2++];
    53         }
    54     }
    55     for(int i = l; i <= m; i++) bit.add(q[i].z, -1);
    56     for(int i = l; i <= r; i++) q[i] = tq[i];
    57 }
    58 
    59 
    60 int main(){
    61     int t;
    62     scanf("%d", &t);
    63     while(t--){
    64         int n, maxz = 0;
    65         memset(ans, 0, sizeof(ans));
    66         scanf("%d", &n);
    67         for(int i = 0; i < n; i++){
    68             scanf("%d %d %d", &q[i].x, &q[i].y, &q[i].z);
    69             q[i].id = i;
    70             maxz = max(maxz, q[i].z);
    71         }
    72         bit.init(maxz);
    73         sort(q, q + n);
    74         for(int i = n - 2; i >= 0; i--){
    75             if(q[i] == q[i + 1]) ans[q[i].id] = ans[q[i + 1].id] + 1; 
    76         }
    77         CDQ(0, n - 1);
    78         for(int i = 0; i < n; i++){
    79             printf("%d
    ", ans[i]);
    80         }
    81     }
    82 }
    View Code

     一维快排、二维CDQ、三维CDQ、(如果有思维就树状数组)

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 const int maxn = 100010;
     4 
     5 struct Node{
     6     int x, y, z, id;
     7     int fg;
     8     bool operator < (const Node &a)const{
     9         if(x != a.x) return x < a.x;
    10         if(y != a.y) return y < a.y;
    11         return z < a.z;
    12     }
    13     bool operator == (const Node &a)const{
    14         return x == a.x && y == a.y && z == a.z;
    15     }
    16 }q[maxn], tq1[maxn], tq2[maxn];
    17 int res[maxn];
    18 
    19 void CDQ2(int l, int r){
    20     if(l == r) return;
    21     int m = l + r >> 1;
    22     CDQ2(l, m); CDQ2(m + 1, r);
    23     int f1 = l, f2 = m + 1, i = l;
    24     int cnt = 0;
    25     while(f1 <= m || f2 <= r){
    26         if(f2 > r || (f1 <= m && tq1[f1].z <= tq1[f2].z)) {
    27             if(tq1[f1].fg == 0) cnt++;
    28             tq2[i++] = tq1[f1++];
    29         }else{
    30             if(tq1[f2].fg == 1) res[tq1[f2].id] += cnt;
    31             tq2[i++] = tq1[f2++];
    32         }
    33     }
    34     for(int i = l; i <= r; i++) tq1[i] = tq2[i];
    35 }
    36 
    37 void CDQ1(int l, int r){
    38     if(l == r) return;
    39     int m = l + r >> 1;
    40     CDQ1(l, m); CDQ1(m + 1, r);
    41     int f1 = l, f2 = m + 1, i = l;
    42     while(f1 <= m || f2 <= r){
    43         if(f2 > r || (f1 <= m && q[f1].y <= q[f2].y)) {
    44             q[f1].fg = 0;
    45             tq1[i++] = q[f1++];
    46         }else{
    47             q[f2].fg = 1;
    48             tq1[i++] = q[f2++];
    49         }
    50     }
    51     for(int i = l; i <= r; i++) q[i] = tq1[i];
    52     CDQ2(l, r);
    53 }
    54 
    55 int main(){
    56     int t;
    57     scanf("%d", &t);
    58     while(t--){
    59         int n;
    60         memset(res, 0, sizeof res);
    61         scanf("%d", &n);
    62         for(int i = 0; i < n; i++){
    63             scanf("%d %d %d", &q[i].x, &q[i].y, &q[i].z);
    64             q[i].id = i;
    65         }
    66         sort(q, q + n);
    67         for(int i = n - 2; i >= 0; i--){
    68             if(q[i] == q[i + 1]) res[q[i].id] = res[q[i + 1].id] + 1;
    69         }
    70         CDQ1(0, n - 1);
    71         for(int i = 0; i < n; i++) printf("%d
    ", res[i]);
    72     }
    73 }
    View Code
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  • 原文地址:https://www.cnblogs.com/yijiull/p/8340244.html
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