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;
}