UVA 12003 Array Transformer

Array Transformer

Time Limit: 5000ms

Memory Limit: 131072KB

This problem will be judged on UVA. Original ID: 12003
64-bit integer IO format: %lld      Java class name: Main

Write a program to transform an array A[1], A[2],..., A[n] according to m instructions. Each instruction (L, R, v, p) means: First, calculate how many numbers from A[L] to A[R](inclusive) are strictly less than v, call this answer k. Then, change the value of A[p] to u*k/(R - L + 1), here we use integer division (i.e. ignoring fractional part).

Input

The first line of input contains three integer n, m, u ( 1n300, 000, 1m50, 000, 1u1, 000, 000, 000). Each of the next n lines contains an integer A[i] ( 1A[i]u). Each of the next m lines contains an instruction consisting of four integers L, R, v, p ( 1LRn, 1vu, 1pn).

Output

Print n lines, one for each integer, the final array.

Sample Input

10 1 11
1
2
3
4
5
6
7
8
9
10
2 8 6 10

Sample Output

1
2
3
4
5
6
7
8
9
6

Explanation: There is only one instruction: L = 2, R = 8, v = 6, p = 10. There are 4 numbers (2,3,4,5) less than 6, so k = 4. The new number in A[10] is 11*4/(8 - 2 + 1) = 44/7 = 6.

解题：分块

参考lrj同学的那本大白

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn = 300000 + 10;
 5 const int SIZE = 4096;
 6 int n,m,u,A[maxn],block[maxn/SIZE+1][SIZE];
 7 void init() {
 8     scanf("%d%d%d",&n,&m,&u);
 9     int b = 0,j = 0;
10     for(int i = 0; i < n; ++i) {
11         scanf("%d",A+i);
12         block[b][j] = A[i];
13         if(++j == SIZE) {
14             b++;
15             j = 0;
16         }
17     }
18     for(int i = 0; i < b; ++i)
19         sort(block[i],block[i] + SIZE);
20     if(j) sort(block[b],block[b]+j);
21 }
22 int query(int L,int R,int v) {
23     int lb = L/SIZE,rb = R/SIZE,k = 0;
24     if(lb == rb) {
25         for(int i = L; i <= R; ++i)
26             k += (A[i] < v);
27     } else {
28         for(int i = L; i < (lb+1)*SIZE; ++i)
29             if(A[i] < v) ++k;
30         for(int i = rb*SIZE; i <= R; ++i)
31             if(A[i] < v) ++k;
32         for(int i = lb+1; i < rb; ++i)
33             k += lower_bound(block[i],block[i]+SIZE,v) - block[i];
34     }
35     return k;
36 }
37 void update(int p,int x) {
38     if(A[p] == x) return;
39     int old = A[p],pos = 0,*B = &block[p/SIZE][0];
40     A[p] = x;
41     while(B[pos] < old) ++pos;
42     B[pos] = x;
43     while(pos < SIZE-1 && B[pos] > B[pos + 1]) {
44         swap(B[pos],B[pos+1]);
45         ++pos;
46     }
47     while(pos > 0 && B[pos] < B[pos - 1]) {
48         swap(B[pos],B[pos-1]);
49         --pos;
50     }
51 }
52 int main() {
53     init();
54     while(m--) {
55         int L,R,v,p;
56         scanf("%d%d%d%d",&L,&R,&v,&p);
57         --L;
58         --R;
59         --p;
60         int k = query(L,R,v);
61         update(p,(LL)u*k/(R - L + 1));
62     }
63     for(int i = 0; i < n; ++i)
64         printf("%d\n",A[i]);
65     return 0;
66 }