LOJ6285 数列分块入门9（分块 区间众数）题解

题意：给出区间内的最小众数

思路：分块，离散化每个数，开vector记录每个数p出现的位置，这样就能二分出L，R以内p的个数了。众数有一个性质，用mode(a)表示集合a的众数，那么mode(a∪b) ∈ mode(a)∪b 。那么我先预处理出任意两块的众数f[i][j]，这样众数就是f[i][j]和旁边两块数中的其中一个了，直接遍历这些数即可。

block不能开方，开30能过。都靠玄学....

代码：

#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include <iostream>
#include<algorithm>
#include<unordered_map>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e5 + 10;
const int M = maxn * 30;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const int MOD = 1e4 + 7;
struct Block{
    int l, r;
}b[maxn];
int a[maxn], belong[maxn];
int f[10000][10000];   //i~j块的众数是
int num[maxn];
int n, block;
vector<int> vv;
vector<int> pos[maxn];
void init(){
    for(int i = 1; i <= belong[n]; i++){    //暴力计算f数组
        for(int j = 0; j <= vv.size(); j++) num[j] = 0;
        int mode = INF, NUM = 0;
        for(int j = b[i].l; j <= n; j++){
            num[a[j]]++;
            if(num[a[j]] > NUM || (num[a[j]] == NUM && a[j] < mode)){
                mode = a[j];
                NUM = num[a[j]];
            }
            f[i][belong[j]] = mode;
        }
    }
}
int getNum(int l, int r, int v){
    int t = upper_bound(pos[v].begin(), pos[v].end(), r) - lower_bound(pos[v].begin(), pos[v].end(), l);
    return t;
}
int query(int l, int r){
    int bl = belong[l], br = belong[r];
    int ans = INF, NUM = 0;
    if(bl == br){
        for(int i = l; i <= r; i++){
            int tot = getNum(l, r, a[i]);
            if(tot > NUM || (tot == NUM && a[i] < ans)){
                ans = a[i];
                NUM = tot;
            }
        }
    }
    else{
        for(int i = l; i <= b[bl].r; i++){
            int tot = getNum(l, r, a[i]);
            if(tot > NUM || (tot == NUM && a[i] < ans)){
                ans = a[i];
                NUM = tot;
            }
        }
        if(bl + 1 <= br - 1){
            int v = f[bl + 1][br - 1];
            int tot = getNum(l, r, v);
            if(tot > NUM || (tot == NUM && v < ans)){
                ans = v;
                NUM = tot;
            }
        }
        for(int i = b[br].l; i <= r; i++){
            int tot = getNum(l, r, a[i]);
            if(tot > NUM || (tot == NUM && a[i] < ans)){
                ans = a[i];
                NUM = tot;
            }
        }
    }
    return vv[ans - 1];
}

int main(){
    scanf("%d", &n);
    vv.clear();
    for(int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
        vv.push_back(a[i]);
    }

    sort(vv.begin(), vv.end());
    vv.erase(unique(vv.begin(), vv.end()), vv.end());
    for(int i = 1; i <= n; i++){
        a[i] = lower_bound(vv.begin(), vv.end(), a[i]) - vv.begin() + 1;
    }
    for(int i = 0; i <= vv.size(); i++) pos[i].clear();

    block = 30;
    for(int i = 1; i <= n; i++){
        belong[i] = (i - 1) / block + 1;
        pos[a[i]].push_back(i);
    }
    for(int i = 1; i <= belong[n]; i++){
        b[i].l = (i - 1) * block + 1;
        b[i].r = b[i].l + block - 1;
    }
    b[belong[n]].r = n;

    init();
    for(int i = 1; i <= n; i++){
        int l, r;
        scanf("%d%d", &l, &r);
        printf("%d
", query(l, r));
    }
    return 0;
}