Codeforces 961E - Tufurama
思路:
线段树或者分块
遍历 1 - n - 1,求 区间[i + 1, min(a[i], n)]大于等于 i 的个数,累加起来
线段树:
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define ls rt << 1, l, m
#define rs rt << 1 | 1, m + 1, r
#define mem(a, b) memset(a, b, sizeof(a)) const int N = 2e5 + ;
vector<int> vc[N<<];
int a[N];
void build(int rt, int l, int r) {
if (l == r) {
vc[rt].pb(a[l]);
return ;
}
for (int i = l; i <= r; i++) vc[rt].pb(a[i]);
sort(vc[rt].begin(), vc[rt].end());
int m = l + r >> ;
build(ls);
build(rs);
}
int query(int L, int R, int rt, int l, int r) {
if (L > R) return ;
if (L <= l && r <= R) {
return vc[rt].size()-(lower_bound(vc[rt].begin(), vc[rt].end(), L - ) - vc[rt].begin());
}
int ans = ;
int m = l + r >> ;
if (L <= m) ans += query(L, R, ls);
if (R > m) ans += query(L, R, rs);
return ans;
}
int main() {
int n;
scanf("%d", &n);
for (int i = ; i <= n; i++) scanf("%d", &a[i]);
build(, , n);
LL ans = ;
for (int i = ; i < n; i++) {
ans += query(i + , min(a[i], n), , , n);
}
printf("%lld\n",ans);
return ;
}
分块:
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define mem(a, b) memset(a, b, sizeof(a)) const int N = 2e5 + ;
int a[N], block[], belong[N];
int blo;
vector<int> vc[];
int query(int L, int R){
if (R < L) return ;
int ans = ;
if (belong[L] == belong[R]) {
for (int i = L; i <= R; i++) {
if (a[i] >= L - ) ans++;
}
return ans;
}
for (int i = L; i <= belong[L] * blo; i++){
if (a[i] >= L - ) ans++;
}
for (int i = belong[L] + ; i <= belong[R] - ; i++) {
ans += vc[i].size() - (lower_bound(vc[i].begin(), vc[i].end(), L - ) - vc[i].begin());
}
for (int i = (belong[R] - ) * blo + ; i <= R; i++) {
if (a[i] >= L - ) ans++;
}
return ans;
}
int main() {
int n;
scanf("%d", &n);
for (int i = ; i <= n; i++) scanf("%d", &a[i]);
blo = sqrt(n);
for (int i = ; i <= n; i++) {
belong[i] = (i - ) / blo + ;
}
for (int i = ; i <= n; i++) {
vc[belong[i]].pb(a[i]);
}
for (int i = ; i <= belong[n]; i++) {
sort(vc[i].begin(), vc[i].end());
}
LL ans = ;
for (int i = ; i <= n - ; i++) {
ans += query(i + , min(n, a[i]));
//cout << ans << endl;
}
printf("%lld\n", ans);
return ;
}
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