【BZOJ 3343】教主的魔法-优快云博客

本文详细解析了BZOJ3343题目的分块算法实现，通过代码示例展示了如何使用分块技术解决区间查询与更新问题。重点介绍了算法思路、代码实现细节及关键优化技巧。

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题目来源：BZOJ 3343

思路：

分块

代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 1000010;
int n, m, v[maxn], w[maxn], SZ, l[maxn], r[maxn], bl[maxn];
int a, b, c, cnt, lazy[maxn];
char ch[5];
int calc(int ll, int rr, int val){
    int l = ll, r = rr, ans = rr+1;
    while(l <= r){
        int mid = (l+r)>>1;
        if(w[mid] < val) l = mid + 1;
        else ans = mid, r = mid - 1;
    }
    return rr - ans + 1;
}
int main(){
    scanf("%d%d", &n, &m), SZ = sqrt(n);
    for(int i = 1; i <= n; i ++) scanf("%d", &v[i]), w[i] = v[i];
    for(int i = 1; i <= n; i ++) bl[i] =(i-1)/SZ + 1;
    l[++cnt] = 1;
    for(int i = 2; i <= n+1; i ++){
        if(bl[i] != bl[i-1]) r[cnt] = i-1, l[++ cnt] = i;
    } 
    cnt --;
    for(int i = 1; i <= cnt; i ++) sort(w+l[i], w+r[i]+1);
    for(int i = 1; i <= m; i ++) {
        scanf("%s%d%d%d", ch+1, &a, &b, &c);
        if(ch[1] == 'A'){
            int ans = 0;
            for(int i = bl[a]; i <= bl[b]; i ++){
                if(i == bl[a] && a != l[i]){
                    for(int j = a; j <= r[i]; j ++){
                        if(v[j]+lazy[i] >= c) ans ++;
                    }
                }else if(i == bl[b] && b != r[i]){
                    for(int j = l[i]; j <= b; j ++){
                        if(v[j]+lazy[i] >= c) ans ++;
                    }
                }else{
                    ans += calc(l[i], r[i], c-lazy[i]);
                }
            }
            printf("%d\n", ans);
        }else{
            for(int i = bl[a]; i <= bl[b]; i ++){
                if(i == bl[a] && a != l[i]){
                    for(int j = l[i]; j <= r[i]; j ++) v[j] += lazy[i];
                    lazy[i] = 0;
                    for(int j = a; j <= r[i]; j ++) v[j] += c;
                    for(int j = l[i]; j <= r[i]; j ++) w[j] = v[j];
                    sort(w+l[i], w+r[i]+1);
                }else if(i == bl[b] && b != r[i]){
                    for(int j = l[i]; j <= r[i]; j ++) v[j] += lazy[i];
                    lazy[i] = 0;
                    for(int j = l[i]; j <= b; j ++) v[j] += c;
                    for(int j = l[i]; j <= r[i]; j ++) w[j] = v[j];
                    sort(w+l[i], w+r[i]+1);
                }else{
                    lazy[i] += c;
                }
            }
        }
    }
    return 0;
}