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

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 300000 + 10;
const int SIZE = 4096;
int n,m,u,A[maxn],block[maxn/SIZE+1][SIZE];
void init() {
    scanf("%d%d%d",&n,&m,&u);
    int b = 0,j = 0;
    for(int i = 0; i < n; ++i) {
        scanf("%d",A+i);
        block[b][j] = A[i];
        if(++j == SIZE) {
            b++;
            j = 0;
        }
    }
    for(int i = 0; i < b; ++i)
        sort(block[i],block[i] + SIZE);
    if(j) sort(block[b],block[b]+j);
}
int query(int L,int R,int v) {
    int lb = L/SIZE,rb = R/SIZE,k = 0;
    if(lb == rb) {
        for(int i = L; i <= R; ++i)
            k += (A[i] < v);
    } else {
        for(int i = L; i < (lb+1)*SIZE; ++i)
            if(A[i] < v) ++k;
        for(int i = rb*SIZE; i <= R; ++i)
            if(A[i] < v) ++k;
        for(int i = lb+1; i < rb; ++i)
            k += lower_bound(block[i],block[i]+SIZE,v) - block[i];
    }
    return k;
}
void update(int p,int x) {
    if(A[p] == x) return;
    int old = A[p],pos = 0,*B = &block[p/SIZE][0];
    A[p] = x;
    while(B[pos] < old) ++pos;
    B[pos] = x;
    while(pos < SIZE-1 && B[pos] > B[pos + 1]) {
        swap(B[pos],B[pos+1]);
        ++pos;
    }
    while(pos > 0 && B[pos] < B[pos - 1]) {
        swap(B[pos],B[pos-1]);
        --pos;
    }
}
int main() {
    init();
    while(m--) {
        int L,R,v,p;
        scanf("%d%d%d%d",&L,&R,&v,&p);
        --L;
        --R;
        --p;
        int k = query(L,R,v);
        update(p,(LL)u*k/(R - L + 1));
    }
    for(int i = 0; i < n; ++i)
        printf("%d
",A[i]);
    return 0;
}