Super Mario HDU - 4417 （主席树）

Posted 2021-01-27 h-riven

Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.

InputThe first line follows an integer T, the number of test data. 
For each test data: 
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries. 
Next line contains n integers, the height of each brick, the range is [0, 1000000000]. 
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)OutputFor each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query. 
Sample Input

1
10 10
0 5 2 7 5 4 3 8 7 7 
2 8 6
3 5 0
1 3 1
1 9 4
0 1 0
3 5 5
5 5 1
4 6 3
1 5 7
5 7 3

Sample Output

Case 1:
4
0
0
3
1
2
0
1
5
1

给出 n 个数，m个询问，对于每个询问，求出在区间[l, r]内小于等于 h 的数的个数。
主席树的模板题，数很大，所以需要离散化一下，维护前缀个数和。
查询的时候先找到第一个小于等于 h 的离散化后对应的下标，利用主席树找区间内值小于等于 k 的数的和。

  1 /*
  2           .
  3          ‘;;;;;.
  4         ‘!;;;;;;!;`
  5        ‘!;|&#@|;;;;!:
  6       `;;!&####@|;;;;!:
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  8      ‘!;;;;;;;;[email protected]###&|;;|%!;!$|;;;;|&&;.
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 10      ;!;;!!;;;;;|$$&@##$;;;::‘‘‘‘‘::;;;;|&|%@$|;;;;;;;;;;;;;;;;!$;
 11      ;|;;;;;;;;;;;;;;;;;;!%@#####&!:::;!;;;;;;;;;;!&####@%!;;;;$%`
 12     `!!;;;;;;;;;;!|%%|!!;::;;|@##%|$|;;;;;;;;;;;;!|%$#####%;;;%&;
 13     :@###&!:;;!!||%%%%%|!;;;;;||;;;;||!$&&@@%;;;;;;;|$$##$;;;%@|
 14     ;|::;;;;;;;;;;;;|&&$|;;[email protected]&$!;;;;!;;;;;;;;;;;;;;;;!%|;;;%@%.
 15    `!!;;;;;;;!!!!;;;;;[email protected]@@&&&&&@$!;!%|;;;;!||!;;;;;!|%%%!;;%@|.
 16 %&&$!;;;;;!;;;;;;;;;;;|$&&&&&&&&&@@%!%%;!||!;;;;;;;;;;;;;$##!
 17 !%;;;;;;!%!:;;;;;;;;;;!$&&&&&&&&&&@##&%|||;;;!!||!;;;;;;;$&:
 18 ‘:|@###%;:;;;;;;;;;;;;!%$&&&&&&@@$!;;;;;;;!!!;;;;;%&!;;|&%.
 19  [email protected]|;;;;;;;;;;;;;;;;;;|%|$&&$%&&|;;;;;;;;;;;;!;;;;;!&@@&‘
 20   .:%#&!;;;;;;;;;;;;;;!%|$$%%&@%;;;;;;;;;;;;;;;;;;;!&@:
 21   .%$;;;;;;;;;;;;;;;;;;|[email protected]&|;;;;;;;;;;;;;;;;;;;;%@%.
 22     !&!;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;|@#;
 23      `%$!;;;;;;;;;;;[email protected]|;;;;;;;;;;;;;;;;;;;;;;;;!%[email protected]#@|.
 24        .|@%!;;;;;;;;;!$&%||;;;;;;;;;;;;;;;;;!%[email protected]#|.
 25            ;&$!;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;%#####|.
 26            |##$|!;;;;;;::‘‘:;;;;;;;;;;;;;!%[email protected]#@;
 27           ;@&|;;;;;;;::‘‘‘‘‘‘:;;;;;;;|$&@###@|`
 28         .%##@|;;;;:::‘‘‘‘‘‘‘‘‘‘::;!%&##$‘
 29       `$##@[email protected]@&|!!;;;:‘‘‘‘‘::::;;;;;|&#%.
 30     ;&@##&$%!;;;;;;::‘‘‘‘‘‘‘‘::;!|%[email protected]#@&@@:
 31      .%@&$$|;;;;;;;;;;:‘‘‘‘:‘‘‘‘::;;;%@#@@#%.
 32     :@##@###@$$$$$|;;:‘‘‘‘:;;!!;;;;;;!$#@@#$;`
 33      `%@$$|;;;;;;;;:‘‘‘‘‘‘‘::;;;;|%$$|!!&###&‘
 34      |##&%!;;;;;::‘‘‘‘‘‘‘‘‘‘‘‘::;;;;;;;[email protected]&:`!‘
 35     :;[email protected]$|;;;;;;;::‘‘‘‘‘‘‘‘‘‘‘:;;;;;;;;!%&@$:                 [email protected]#$‘
 36       |##@@&%;;;;;::‘‘‘‘‘‘‘‘‘:;;;;;;;!%&@#@$%:              ‘%%!%&;
 37       |&%!;;;;;;;%$!:‘‘‘‘‘‘‘:|%!;;;;;;;;|&@%||`            ‘%$|!%&;
 38       |@%!;;!!;;;||;:‘‘‘‘‘‘:;%$!;;;;!%%%&#&%$&:           .|%;:!&%`
 39       [email protected]&%;;;;;;;||;;;:‘‘::;;%$!;;;;;;;|&@%;!$;          `%&%!!$&:
 40       ‘$$|;!!!!;;||;;;;;;;;;;%%;;;;;;;|@@|!$##;         !$!;:!$&:
 41        |#&|;;;;;;!||;;;;;;;;!%|;;;;!$##$;;;;|%‘      `%$|%%;|&$‘
 42         |&%!;;;;;;|%;;;;;;;;$$;;;;;;|&&|!|%&&;  .:%&$!;;;:[email protected]!
 43         `%#&%!!;;;;||;;;;;!$&|;;;!%%%@&!;;;!!;;;|%!;;%@$!%@!
 44         !&!;;;;;;;;;||;;%&!;;;;;;;;;%@&!;;!&$;;;|&%;;;%@%`
 45        ‘%|;;;;;;;;!!|$|%&%;;;;;;;;;;|&#&|!!||!!|%[email protected]@|‘
 46        .!%%&%‘`|$;       :|$#%|@#&;%#%.
 47 */
 48 #include <map>
 49 #include <set>
 50 #include <list>
 51 #include <ctime>
 52 #include <cmath>
 53 #include <stack>
 54 #include <queue>
 55 #include <string>
 56 #include <vector>
 57 #include <cstdio>
 58 #include <bitset>
 59 #include <cstdlib>
 60 #include <cstring>
 61 #include <iostream>
 62 #include <algorithm>
 63 #define  lowbit(x)  x & (-x)
 64 #define  mes(a, b)  memset(a, b, sizeof a)
 65 #define  fi         first
 66 #define  se         second
 67 #define  pii        pair<int, int>
 68 #define  INOPEN     freopen("in.txt", "r", stdin)
 69 #define  OUTOPEN    freopen("out.txt", "w", stdout)
 70 
 71 typedef unsigned long long int  ull;
 72 typedef long long int ll;
 73 const int    maxn = 1e5 + 10;
 74 const int    maxm = 1e5 + 10;
 75 const int    mod  = 1e9 + 7;
 76 const ll     INF  = 1e18 + 100;
 77 const int    inf  = 0x3f3f3f3f;
 78 const double pi   = acos(-1.0);
 79 const double eps  = 1e-8;
 80 using namespace std;
 81 
 82 int n, m;
 83 int cas, tol, T;
 84 
 85 struct Node {
 86     int l, r;
 87     int sum;
 88 } node[maxn * 20];
 89 int a[maxn];
 90 int rt[maxn];
 91 vector<int> vv;
 92 
 93 void init() {
 94     tol = 0;
 95     mes(rt, 0);
 96     vv.clear();
 97 }
 98 
 99 int getid(int x) {
100     return lower_bound(vv.begin(), vv.end(), x) - vv.begin() + 1;
101 }
102 
103 void update(int l, int r, int &x, int y, int pos) {
104     tol++;
105     node[tol] = node[y];
106     node[tol].sum ++;
107     x = tol;
108     if(l == r)  return ;
109     int mid = (l + r) >> 1;
110     if(pos <= mid)
111         update(l, mid, node[x].l, node[y].l, pos);
112     else
113         update(mid+1, r, node[x].r, node[y].r, pos);
114 }
115 
116 int query(int l, int r, int x, int y, int pos) {
117     if(l == r) {
118         if(l <= pos)
119             return node[y].sum - node[x].sum;
120         else
121             return 0;
122     }
123     if(r <= pos)
124         return node[y].sum - node[x].sum;
125     int mid = (l +r) >> 1;
126     if(pos <= mid)
127         return query(l, mid, node[x].l, node[y].l, pos);
128     else {
129         int tmp = node[node[y].l].sum - node[node[x].l].sum;
130         return tmp + query(mid+1, r, node[x].r, node[y].r, pos);
131     }
132 }
133 
134 int main() {
135     cas = 1;
136     scanf("%d", &T);
137     while(T--) {
138         init();
139         scanf("%d%d", &n, &m);
140         for(int i=1; i<=n; i++) {
141             scanf("%d", &a[i]);
142             vv.push_back(a[i]);
143         }
144         sort(vv.begin(), vv.end());
145         vv.erase(unique(vv.begin(), vv.end()), vv.end());
146         for(int i=1; i<=n; i++) {
147             int id = getid(a[i]);
148             update(1, n, rt[i], rt[i-1], id);
149         }
150         printf("Case %d:
", cas++);
151         while(m--) {
152             int u, v, d;
153             scanf("%d%d%d", &u, &v, &d);
154             u++, v++;
155             int id = getid(d);
156             if(vv[id-1] != d)
157                 id--;
158             int ans = query(1, n, rt[u-1], rt[v], id);
159             printf("%d
", ans);
160         }
161     }
162     return 0;
163 }

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