uva12003 - Array Transformer 分块结构

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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 lessthan v, call this answer k. Then, change the value of A[p] to u ∗ k/(R − L + 1), here we use integerdivision (i.e. ignoring fractional part).InputThe first line of input contains three integer n, m, u (1 ≤ n ≤ 300, 000, 1 ≤ m ≤ 50, 000, 1 ≤ u ≤1, 000, 000, 000). Each of the next n lines contains an integer A[i] (1 ≤ A[i] ≤ u). Each of the nextm lines contains an instruction consisting of four integers L, R, v, p (1 ≤ L ≤ R ≤ n, 1 ≤ v ≤ u,1 ≤ p ≤ n).OutputPrint n lines, one for each integer, the final array.Sample Input10 1 11123456789102 8 6 10Sample Output1234567896Explanation: 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.

输入N（1<=N<=300000),M(1<=M<=50000),U(1<=U<=10^9)，接下来输入数组a（N个数），M次操作，每次操作输入L,R,V,P，计算出[L,R]区间内小于V的数的个数k，把A[P]修改为U*k/(R-L+1)。最后输出M次操作后的数组。

把数组分块，每块大小在根号N附近（这样比较快），对每块排序，这样每次询问L,R的时候在L,R所在的块进行暴力查找，L,R之间的块用二分查找。修改的时候不光要修改原数组，还要修改分块数组，找到P所在的块，然后找到要修改的值所在的位置，把这个位置替换为新值，再维护块内的顺序。找L，R所在块的位置很好找，因为是按顺序分的，所以直接用位置除以块大小就是块编号。

分块法其实也就是划分成小范围的暴力，这里如果块大小取根号N的话，复杂度O(M*√N*log√N)，白书上说分块法非常有用，很多用伸展树或树套树能解决的问题用分块法。

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;const int MAXN=300010;const int SIZE=4096;const int INF=0x3f3f3f3f;int N,M,U,L,R,V,P;int a[MAXN],block[MAXN/SIZE+1][SIZE];int query(int l,int r,int v){    int lb=l/SIZE,rb=r/SIZE;    int ret=0;    if(lb==rb){        for(int i=l;i<=r;i++) if(a[i]<v) ret++;    }    else{        for(int i=l;i<(lb+1)*SIZE;i++) if(a[i]<v) ret++;        for(int i=rb*SIZE;i<=r;i++) if(a[i]<v) ret++;        for(int i=lb+1;i<rb;i++) ret+=lower_bound(block[i],block[i]+SIZE,v)-block[i];    }    return ret;}void change(int p,int x){    if(a[p]==x) return;    int old=a[p],pos=0,*b=block[p/SIZE];    a[p]=x;    while(b[pos]<old) pos++;    b[pos]=x;    if(x>old){        while(pos<SIZE-1&&b[pos]>b[pos+1]){            swap(b[pos+1],b[pos]);            pos++;        }    }    else{        while(pos>0&&b[pos]<b[pos-1]){            swap(b[pos-1],b[pos]);            pos--;        }    }}int main(){    freopen("in.txt","r",stdin);    while(scanf("%d%d%d",&N,&M,&U)!=EOF){        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);        while(M--){            scanf("%d%d%d%d",&L,&R,&V,&P);            L--;            R--;            P--;            int k=query(L,R,V);            change(P,(LL)U*k/(R-L+1));        }        for(int i=0;i<N;i++) printf("%d\n",a[i]);    }    return 0;}

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