【题目大意】
有$n$个位置，每个位置有一个数$x_i$，代表从$i$经过1步可以到达的点在$[\max(1, i-x_i), \min(i+x_i, n)]$中。
定义$(i,j)$的距离表示从$i$到$j$经过多少步，从$j$到$i$经过多少步，这两个取最小值。
求任意两点间最大的距离。
$1\leq n \leq 10^5, 1 \leq x_i < n$
【题解】
每个点经过若干次能过到达的，显然是一个区间。
考虑倍增，$[L_{x,i}, R_{x,i}]$表示从$x$开始，经过$2^i$步，到达的区间。
这个可以通过倍增+线段树来解决，线段树维护最小值和最大值，对应区间两个端点。
对于询问，我们二分答案后，问题转化与是否能找出一个点对步数$> x$（$x$为我们二分的值）
通过倍增我们可以求出每个点经过$x$步后到达的区间，设为$[L', R']$。那么不能到达的就是$[1, L']$和$[R',n]$。
问题转化为，$[1, L']$和$[R', n]$是否可以经过$x$步以内到达$i$这个点。
这个可以通过记录一个前缀和以及一个后缀和来解决。
这里的“和”指的是区间交。
然后就行啦！
时间复杂度$O(nlog^3n)$
upd: 线段树有点问题，已经修正
=========================分割线=============================
upd: 之前的复杂度有点问题，是$O(nlog^3n)$的，理论上是过不去的，但是常数太优秀了（逃
我们可以把二分改为从高位往低位确定答案的每位，每次就不需要倍增了，就在之前的基础上，直接做即可。这样就是$O(nlog^2n)$的了
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm> using namespace std; typedef long long ll;
typedef unsigned long long ull;
typedef long double ld; const int N = 1e5 + , M = 2e5 + ;
const int inf = 1e9; # define bit(x, i) (((x) >> (i)) & ) int n;
struct pa {
int x, y;
pa() {}
pa(int x, int y) : x(x), y(y) {}
friend pa operator + (pa a, pa b) {
return pa(min(a.x, b.x), max(a.y, b.y));
}
friend pa operator - (pa a, pa b) {
return pa(max(a.x, b.x), min(a.y, b.y));
}
}; inline bool out(int x, pa t) {
return x < t.x || x > t.y;
} int L[][N], R[][N]; struct SMT {
pa w[M << ];
# define ls (x<<)
# define rs (x<<|)
inline void set(int n) {
for (int i=; i<=n+n; ++i) w[i] = pa(inf, -inf);
}
inline void build(int x, int l, int r, int p) {
if(l == r) {
w[x] = pa(L[p][l], R[p][l]);
return ;
}
int mid = l+r>>;
build(ls, l, mid, p);
build(rs, mid+, r, p);
w[x] = w[ls] + w[rs];
// printf("p = %d, [%d, %d], [%d, %d]\n", p, l, r, w[x].x, w[x].y);
}
inline pa query(int x, int l, int r, int L, int R) {
if(L <= l && r <= R) return w[x];
int mid = l+r>>;
if(L > mid) return query(rs, mid+, r, L, R);
else if (R <= mid) return query(ls, l, mid, L, R);
else return query(ls, l, mid, L, mid) + query(rs, mid+, r, mid+, R);
}
# undef ls
# undef rs
}T[]; // have dis > x
pa c[N];
pa t[N];
pa f[N], g[N]; inline bool chk(int x) {
pa tmp;
for (int i=; i<=n; ++i) {
tmp.x = t[i].x, tmp.y = t[i].y;
tmp = T[x].query(, , n, tmp.x, tmp.y);
c[i] = tmp;
// printf(" i = %d, [%d, %d]\n", i, c[i].x, c[i].y);
}
f[] = c[]; g[n] = c[n];
for (int i=; i<=n; ++i) f[i] = f[i-] - c[i];
for (int i=n-; i; --i) g[i] = g[i+] - c[i];
for (int i=; i<=n; ++i) {
if(c[i].x == && c[i].y == n) continue;
// [1, c[i].x-1], [c[i].y+1], n]
if(c[i].x != ) {
if(out(i, f[c[i].x-])) return true;
}
if(c[i].y != n) {
if(out(i, g[c[i].y+])) return true;
}
}
return false;
}
// # include <time.h>
int main() {
freopen("jump.in", "r", stdin);
freopen("jump.out", "w", stdout);
cin >> n;
for (int i=; i<=; ++i) T[i].set(n);
for (int i=, x; i<=n; ++i) {
scanf("%d", &x);
L[][i] = max(i-x, );
R[][i] = min(i+x, n);
}
pa tem;
for (int j=; j<=; ++j) {
T[j-].build(, , n, j-);
for (int i=; i<=n; ++i) {
tem = T[j-].query(, , n, L[j-][i], R[j-][i]);
L[j][i] = tem.x, R[j][i] = tem.y;
}
}
T[].build(, , n, ); // for (int j=0; j<=5; ++j)
// for (int i=1; i<=n; ++i) printf("%d %d L = %d, R = %d\n", i, j, L[j][i], R[j][i]); int l = , r = n-, mid, ans = ;
for (int i=; i<=n; ++i) t[i] = pa(i, i);
for (int i=; ~i; --i) {
if(chk(i)) {
for (int j=; j<=n; ++j)
t[j] = c[j];
ans += (<<i);
}
} cout << ans + << endl;
// cerr << clock() << " ms\n";
return ;
}
/*
8
7 1 1 1 1 1 1 7 10
2 2 1 2 2 1 2 2 1 2
*/
下面是$O(nlog^3n)$的代码
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm> using namespace std; typedef long long ll;
typedef unsigned long long ull;
typedef long double ld; const int N = 1e5 + , M = 2e5 + ;
const int inf = 1e9; # define bit(x, i) (((x) >> (i)) & ) int n;
struct pa {
int x, y;
pa() {}
pa(int x, int y) : x(x), y(y) {}
friend pa operator + (pa a, pa b) {
return pa(min(a.x, b.x), max(a.y, b.y));
}
friend pa operator - (pa a, pa b) {
return pa(max(a.x, b.x), min(a.y, b.y));
}
}; inline bool out(int x, pa t) {
return x < t.x || x > t.y;
} int L[][N], R[][N]; struct SMT {
pa w[M << ];
# define ls (x<<)
# define rs (x<<|)
inline void set(int n) {
for (int i=; i<=n+n; ++i) w[i] = pa(inf, -inf);
}
inline void build(int x, int l, int r, int p) {
if(l == r) {
w[x] = pa(L[p][l], R[p][l]);
return ;
}
int mid = l+r>>;
build(ls, l, mid, p);
build(rs, mid+, r, p);
w[x] = w[ls] + w[rs];
// printf("p = %d, [%d, %d], [%d, %d]\n", p, l, r, w[x].x, w[x].y);
}
inline pa query(int x, int l, int r, int L, int R) {
if(L <= l && r <= R) return w[x];
int mid = l+r>>;
if(L > mid) return query(rs, mid+, r, L, R);
else if (R <= mid) return query(ls, l, mid, L, R);
else return query(ls, l, mid, L, mid) + query(rs, mid+, r, mid+, R);
}
# undef ls
# undef rs
}T[]; // have dis > x
pa c[N]; pa f[N], g[N]; inline bool chk(int x) {
// cout << "x = " << x << endl;
pa t;
for (int i=; i<=n; ++i) {
t.x = t.y = i;
for (int j=; ~j; --j)
if(bit(x, j)) t = T[j].query(, , n, t.x, t.y);
c[i] = t;
// printf(" i = %d, [%d, %d]\n", i, c[i].x, c[i].y);
}
f[] = c[]; g[n] = c[n];
for (int i=; i<=n; ++i) f[i] = f[i-] - c[i];
for (int i=n-; i; --i) g[i] = g[i+] - c[i];
for (int i=; i<=n; ++i) {
if(c[i].x == && c[i].y == n) continue;
// [1, c[i].x-1], [c[i].y+1], n]
if(c[i].x != ) {
if(out(i, f[c[i].x-])) return true;
}
if(c[i].y != n) {
if(out(i, g[c[i].y+])) return true;
}
}
return false;
}
// # include <time.h>
int main() {
freopen("jump.in", "r", stdin);
freopen("jump.out", "w", stdout);
cin >> n;
for (int i=; i<=; ++i) T[i].set(n);
for (int i=, x; i<=n; ++i) {
scanf("%d", &x);
L[][i] = max(i-x, );
R[][i] = min(i+x, n);
}
pa tem;
for (int j=; j<=; ++j) {
T[j-].build(, , n, j-);
for (int i=; i<=n; ++i) {
tem = T[j-].query(, , n, L[j-][i], R[j-][i]);
L[j][i] = tem.x, R[j][i] = tem.y;
}
}
T[].build(, , n, ); // for (int j=0; j<=5; ++j)
// for (int i=1; i<=n; ++i) printf("%d %d L = %d, R = %d\n", i, j, L[j][i], R[j][i]); int l = , r = n-, mid, ans;
while() {
if(r-l <= ) {
for (int i=r; i>=l; --i)
if(chk(i)) {
ans = i;
break;
}
break;
}
mid = l+r>>;
if(chk(mid)) l = mid;
else r = mid;
} cout << ans + << endl;
// cerr << clock() << " ms\n";
return ;
}
/*
8
7 1 1 1 1 1 1 7 10
2 2 1 2 2 1 2 2 1 2
*/