[学习笔记]启发式分治

启发式分治

给定n个数，求满足某种条件的点对数目或最大权值，而这个最大权值与点对(a,b)的区间[a,b]的区间最大/最小值有关。
那么这时就可以考虑分治，对于区间[L,R]，找到最小/大值所在位置，然后处理横跨最小/大值所在位置的点对，然后递归处理子区间。

HUD-Make Rounddog Happy

• 求有多少区间，满足其中每个数都不同，并且区间最大值大于k
• 预处理区间最大值，以及每个数往左往右最远不相同的位置
• 暴力处理区间小的，然后递归

#pragma GCC optimize(2)

#include <bits/stdc++.h>

#define ll long long
using namespace std;
const int maxn = 3e5 + 7;
int stmax[maxn][20], pos[maxn][20], poww[20], logg[maxn];
int n, s[maxn], p, L[maxn], R[maxn];
bool vis[maxn];

void get_st() {
    poww[0] = 1;
    for (int i = 1; i < 20; ++i) poww[i] = poww[i - 1] << 1;
    for (int i = 2; i <= n; ++i) logg[i] = logg[i >> 1] + 1;
    int temp = 1;
    for (int j = 1; j <= logg[n]; ++j) {
        for (int i = 1; i <= n - temp * 2 + 1; ++i) {
            if (stmax[i][j - 1] >= stmax[i + temp][j - 1]) {
                stmax[i][j] = stmax[i][j - 1];
                pos[i][j] = pos[i][j - 1];
            } else {
                stmax[i][j] = stmax[i + temp][j - 1];
                pos[i][j] = pos[i + temp][j - 1];
            }
        }
        temp <<= 1;
    }
}

int query_pos(int l, int r) {
    int k = logg[r - l + 1];
    if (stmax[l][k] >= stmax[r - poww[k] + 1][k]) {
        return pos[l][k];
    } else {
        return pos[r - poww[k] + 1][k];
    }
}

void get_diff() {
    vis[s[1]] = 1;
    int r = 2;
    for (int i = 1; i <= n; ++i) {
        while (r <= n && !vis[s[r]]) {
            vis[s[r]] = 1;
            r++;
        }
        vis[s[i]] = 0;
        R[i] = r - 1;
    }
    vis[s[n]] = 1;
    r = n - 1;
    for (int i = n; i >= 1; --i) {
        while (r >= 1 && !vis[s[r]]) {
            vis[s[r]] = 1;
            r--;
        }
        vis[s[i]] = 0;
        L[i] = r + 1;
    }
}

ll ans = 0;

void solve(int l, int r) {
    if (l > r) return;
    int mid = query_pos(l, r);
    if (r - mid > mid - l) {
        for (int i = l; i <= mid; ++i) {
            int d = max(mid, s[mid] - p + i - 1);
            int dd = min(R[i], r);
            if (dd < d)continue;
            ans += dd - d + 1;
        }
    } else {
        for (int i = r; i >= mid; --i) {
            int d = min(mid, p + 1 + i - s[mid]);
            int dd = max(L[i], l);
            if (dd > d)continue;
            ans += d - dd + 1;
        }
    }
    solve(l, mid - 1);
    solve(mid + 1, r);
}

int main() {
    int _;
    scanf("%d", &_);
    while (_--) {
        scanf("%d%d", &n, &p);
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &s[i]);
            stmax[i][0] = s[i];
            pos[i][0] = i;
            vis[i] = 0;
        }
        get_st();
        get_diff();
        ans = 0;
        solve(1, n);
        printf("%lld\n", ans);
    }
    return 0;
}

Removing Stones

• 题意最后可以化为求有多少个区间，满足区间最大值小于区间和的两倍

• 最大值还是ST表求，然后二分求满足区间和的位置

#include <bits/stdc++.h>
 
#define ll long long
using namespace std;
const int maxn = 3e5 + 7;
int stmax[maxn][20], pos[maxn][20], poww[20], logg[maxn];
int n, s[maxn];
ll sum[maxn],sum1[maxn];
 
void get_st() {
    poww[0] = 1;
    for (int i = 1; i < 20; ++i) poww[i] = poww[i - 1] << 1;
    for (int i = 2; i <= n; ++i) logg[i] = logg[i >> 1] + 1;
    int temp = 1;
    for (int j = 1; j <= logg[n]; ++j) {
        for (int i = 1; i <= n - temp * 2 + 1; ++i) {
            if (stmax[i][j - 1] >= stmax[i + temp][j - 1]) {
                stmax[i][j] = stmax[i][j - 1];
                pos[i][j] = pos[i][j - 1];
            } else {
                stmax[i][j] = stmax[i + temp][j - 1];
                pos[i][j] = pos[i + temp][j - 1];
            }
        }
        temp <<= 1;
    }
}
 
int query_pos(int l, int r) {
    int k = logg[r - l + 1];
    if (stmax[l][k] >= stmax[r - poww[k] + 1][k]) {
        return pos[l][k];
    } else {
        return pos[r - poww[k] + 1][k];
    }
}
 
ll ans = 0;
int erfen1(int l,int r,ll x){
    while(l<r){
        int mid=(l+r)>>1;
        if(sum[mid]>=x) r=mid;else l=mid+1;
    }
    return l;
}
int erfen2(int l,int r,ll x){
    while(l<r){
        int mid=(l+r+1)>>1;
        if(sum1[mid]<=x)l=mid; else r=mid-1;
    }
    return l;
}
void solve(int l, int r) {
    if (l >= r)return;
    int mid = query_pos(l, r);
    if (mid - l < r - mid) {
        for (int i = l; i <= mid; ++i) {
            int pp=erfen1(mid,r+1,s[mid]*2ll+sum1[i]);
            //cout<<pp<<" "<<s[mid]*2ll+sum1[i]<<" "<<mid<<endl;
            ans+=r-pp+1;
        }
    } else {
        for (int i = r; i >= mid; --i) {
            int pp=erfen2(l-1,mid,sum[i]-s[mid]*2ll);
            ans+=pp-l+1;
        }
    }
    solve(l, mid - 1);
    solve(mid + 1, r);
}
 
int main() {
    int _;
    scanf("%d", &_);
    while (_--) {
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &s[i]);
            stmax[i][0] = s[i];
            pos[i][0] = i;
            sum[i]=sum[i-1]+s[i];
            sum1[i]=sum[i-1];
        }
        get_st();
        ans=0;
        solve(1,n);
        printf("%lld\n",ans);
    }
    return 0;
}