「6月雅礼集训 2017 Day11」jump

【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。

$1leq n leq 10^5, 1 leq x_i < n$

【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 + 10, M = 2e5 + 10;
const int inf = 1e9;

# define bit(x, i) (((x) >> (i)) & 1)

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[18][N], R[18][N];

struct SMT {
    pa w[M << 1];
    # define ls (x<<1)
    # define rs (x<<1|1)
    inline void set(int n) {
        for (int i=0; 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>>1;
        build(ls, l, mid, p);
        build(rs, mid+1, r, p);
        w[x] = w[ls] + w[rs];
//        printf("p = %d, [%d, %d],  [%d, %d]
", 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>>1;
        if(L > mid) return query(rs, mid+1, 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+1, r, mid+1, R);
    }
    # undef ls
    # undef rs
}T[18];

// have dis > x
pa c[N];
pa t[N];
pa f[N], g[N];

inline bool chk(int x) {
    pa tmp;
    for (int i=1; i<=n; ++i) {
        tmp.x = t[i].x, tmp.y = t[i].y;
        tmp = T[x].query(1, 1, n, tmp.x, tmp.y);
        c[i] = tmp;
//        printf("  i = %d, [%d, %d]
", i, c[i].x, c[i].y);
    }
    f[1] = c[1]; g[n] = c[n];
    for (int i=2; i<=n; ++i) f[i] = f[i-1] - c[i];
    for (int i=n-1; i; --i) g[i] = g[i+1] - c[i];
    for (int i=1; i<=n; ++i) {
        if(c[i].x == 1 && c[i].y == n) continue;
        // [1, c[i].x-1],  [c[i].y+1], n]
        if(c[i].x != 1) {
            if(out(i, f[c[i].x-1])) return true;
        }
        if(c[i].y != n) {
            if(out(i, g[c[i].y+1])) return true;    
        }
    }
    return false;
}
// # include <time.h>
int main() {
    freopen("jump.in", "r", stdin);
    freopen("jump.out", "w", stdout);
    cin >> n;
    for (int i=0; i<=17; ++i) T[i].set(n);
    for (int i=1, x; i<=n; ++i) {
        scanf("%d", &x);
        L[0][i] = max(i-x, 1);
        R[0][i] = min(i+x, n);
    }
    pa tem;
    for (int j=1; j<=17; ++j) {
        T[j-1].build(1, 1, n, j-1);
        for (int i=1; i<=n; ++i) {
            tem = T[j-1].query(1, 1, n, L[j-1][i], R[j-1][i]);
            L[j][i] = tem.x, R[j][i] = tem.y;
        }
    }
    T[17].build(1, 1, n, 17);
    
//    for (int j=0; j<=5; ++j) 
//        for (int i=1; i<=n; ++i) printf("%d %d L = %d, R = %d
", i, j, L[j][i], R[j][i]);
    
    int l = 0, r = n-1, mid, ans = 0;
    for (int i=1; i<=n; ++i) t[i] = pa(i, i);
    for (int i=16; ~i; --i) {
        if(chk(i)) {
            for (int j=1; j<=n; ++j) 
                t[j] = c[j];
            ans += (1<<i);
        }
    }

    cout << ans + 1 << endl;
//    cerr << clock() << " ms
";        
    return 0;
}
/*
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 + 10, M = 2e5 + 10;
const int inf = 1e9;

# define bit(x, i) (((x) >> (i)) & 1)

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[18][N], R[18][N];

struct SMT {
    pa w[M << 1];
    # define ls (x<<1)
    # define rs (x<<1|1)
    inline void set(int n) {
        for (int i=0; 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>>1;
        build(ls, l, mid, p);
        build(rs, mid+1, r, p);
        w[x] = w[ls] + w[rs];
//        printf("p = %d, [%d, %d],  [%d, %d]
", 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>>1;
        if(L > mid) return query(rs, mid+1, 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+1, r, mid+1, R);
    }
    # undef ls
    # undef rs
}T[18];

// 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=1; i<=n; ++i) {
        t.x = t.y = i;
        for (int j=17; ~j; --j)
            if(bit(x, j)) t = T[j].query(1, 1, n, t.x, t.y);
        c[i] = t;
//        printf("  i = %d, [%d, %d]
", i, c[i].x, c[i].y);
    }
    f[1] = c[1]; g[n] = c[n];
    for (int i=2; i<=n; ++i) f[i] = f[i-1] - c[i];
    for (int i=n-1; i; --i) g[i] = g[i+1] - c[i];
    for (int i=1; i<=n; ++i) {
        if(c[i].x == 1 && c[i].y == n) continue;
        // [1, c[i].x-1],  [c[i].y+1], n]
        if(c[i].x != 1) {
            if(out(i, f[c[i].x-1])) return true;
        }
        if(c[i].y != n) {
            if(out(i, g[c[i].y+1])) return true;    
        }
    }
    return false;
}
// # include <time.h>
int main() {
    freopen("jump.in", "r", stdin);
    freopen("jump.out", "w", stdout);
    cin >> n;
    for (int i=0; i<=17; ++i) T[i].set(n);
    for (int i=1, x; i<=n; ++i) {
        scanf("%d", &x);
        L[0][i] = max(i-x, 1);
        R[0][i] = min(i+x, n);
    }
    pa tem;
    for (int j=1; j<=17; ++j) {
        T[j-1].build(1, 1, n, j-1);
        for (int i=1; i<=n; ++i) {
            tem = T[j-1].query(1, 1, n, L[j-1][i], R[j-1][i]);
            L[j][i] = tem.x, R[j][i] = tem.y;
        }
    }
    T[17].build(1, 1, n, 17);
    
//    for (int j=0; j<=5; ++j) 
//        for (int i=1; i<=n; ++i) printf("%d %d L = %d, R = %d
", i, j, L[j][i], R[j][i]);
    
    int l = 0, r = n-1, mid, ans;
    while(1) {
        if(r-l <= 3) {
            for (int i=r; i>=l; --i) 
                if(chk(i)) {
                    ans = i;
                    break;
                }
            break;
        }
        mid = l+r>>1;
        if(chk(mid)) l = mid;
        else r = mid;
    }

    cout << ans + 1 << endl;
//    cerr << clock() << " ms
";        
    return 0;
}
/*
8
7 1 1 1 1 1 1 7

10
2 2 1 2 2 1 2 2 1 2
*/