CodeForces 13E Holes（分塊處理）

Holes

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

Little Petya likes to play a lot. Most of all he likes to play a game «Holes». This is a game for one person with following rules:

There are N holes located in a single row and numbered from left to right with numbers from 1 to N. Each hole has it's own power (hole number i has the power ai). If you throw a ball into hole i it will immediately jump to hole i + ai, then it will jump out of it and so on. If there is no hole with such number, the ball will just jump out of the row. On each of the M moves the player can perform one of two actions:

• Set the power of the hole a to value b.
• Throw a ball into the hole a and count the number of jumps of a ball before it jump out of the row and also write down the number of the hole from which it jumped out just before leaving the row.

Petya is not good at math, so, as you have already guessed, you are to perform all computations.

Input

The first line contains two integers N and M (1 ≤ N ≤ 105, 1 ≤ M ≤ 105) — the number of holes in a row and the number of moves. The second line contains N positive integers not exceeding N — initial values of holes power. The following M lines describe moves made by Petya. Each of these line can be one of the two types:

• 0 a b
• 1 a

Type 0 means that it is required to set the power of hole a to b, and type 1 means that it is required to throw a ball into the a-th hole. Numbers a and b are positive integers do not exceeding N.

Output

For each move of the type 1 output two space-separated numbers on a separate line — the number of the last hole the ball visited before leaving the row and the number of jumps it made.

Sample test(s)

input

8 5
1 1 1 1 1 2 8 2
1 1
0 1 3
1 1
0 3 4
1 2

output

8 7
8 5
7 3

題意：有1～n個洞，每個洞i都有一個對應的值v[i]，當小球落入洞i中後將彈到第i+v[i]個洞裏，直到彈出所有洞。一共m次操作，0 a b表示將洞a的值修改爲b，1 a表示詢問，要求輸出將小球投入洞a後，小球最後彈入的洞和小球彈的次數。

每次修改操作都需要更新，如果暴力一定超時。所以考慮分塊處理，將n個洞分成根號n個區塊，每次只需要在洞a所在的區塊中更新。cnt數組記錄小球彈出所在區塊需要的次數，last數組記錄小球在所在區塊最後彈入的洞，link數組記錄小球從所在區塊彈入其他區塊時首先彈入的洞。這樣，時間複雜度由O(N²)優化爲O(NlogN)。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

int n,m;
int v[100005];//v[i]：第i個洞的值
int cnt[100005];//cnt[i]：跳出第i個洞所在區塊所需要的步數
int last[100005];//last[i]：跳出第i個洞所在區塊時最後待過的洞
int link[100005];//link[i]：從i所在區塊跳出後到達的洞
int block;//每個區塊中洞的個數

void update(int x, int y)//從第x個洞跳到第y個洞
{
    if(y > n)//跳出去了
    {
        link[x] = n + 1;
        last[x] = x;
        cnt[x] = 1;
    }
    else if(x / block == y / block)//跳在同一個區塊中
    {
        link[x] = link[y];
        last[x] = last[y];
        cnt[x] = cnt[y] + 1;
    }
    else//跳到其他區塊
    {
        link[x] = y;
        last[x] = x;
        cnt[x] = 1;
    }
}

void Search(int x)
{
    int num,ans;
    ans = cnt[x];
    num = last[x];
    while(1)
    {
        x = link[x];
        if(x > n) break;
        num = last[x];
        ans += cnt[x];
    }
    printf("%d %d\n",num,ans);
}

int main()
{
    int choice,x,a,b;
    while(scanf("%d%d",&n,&m) != EOF)
    {
        block = ceil(sqrt(n));//開根後進一法取整
        //cout << block << endl;
        for(int i = 1; i <= n; i++)
            scanf("%d",&v[i]);
        for(int i = n; i > 0; i--)//從後往前更新
            update(i,i + v[i]);
        while(m --)
        {
            scanf("%d",&choice);
            if(choice)
            {
                scanf("%d",&x);
                Search(x);
            }
            else
            {
                scanf("%d%d",&a,&b);
                v[a] = b;
                for(int i = a; i && i / block == a / block; i--)//更新所在區塊即可
                    update(i,i + v[i]);
            }
        }
    }
}