Bounce 弹飞绵羊

Bounce 弹飞绵羊

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2002

分块

将整个大区间分成若干块，每个点维护到下一个块需要跳的次数以及会跳到哪个点（分块要注意细节，区间开闭容易弄乱）。

代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#define B (int(sqrt(n)))
#define N 200000
using namespace std;
struct nod{
    int num;
    int next,cost;
}a[N+5];
int n,m;
int main(void){
    scanf("%d",&n);
    for(int i=0;i<n;++i)
        scanf("%d",&a[i].num);
    for(int i=n-1;i>=0;--i){
        int index=i+a[i].num;
        int R=min(n,((i/B)+1)*B);
        if(index>=R){/**直接if(index/B>i/B)会出现始终不成立的情况导致后面陷入死循环**/
            a[i].cost=1;
            a[i].next=index;
        }else{
            a[i].cost=a[index].cost+1;
            a[i].next=a[index].next;
        }
    }
    scanf("%d",&m);
    while(m--){
        int type;
        scanf("%d",&type);
        if(type==1){
            int index,sum=0;
            scanf("%d",&index);
            while(index<n){
                sum+=a[index].cost;
                index=a[index].next;
            }
            printf("%d
",sum);
        }else if(type==2){
            int index,num;
            scanf("%d%d",&index,&num);
            int block=index/B;
            a[index].num=num;
            for(int i=index;i>=block*B;--i){
                int t=a[i].num+i;
                int R=min(n,(block+1)*B);
                if(t>=R){
                    a[i].cost=1;
                    a[i].next=t;
                }else{
                    a[i].cost=a[t].cost+1;
                    a[i].next=a[t].next;
                }
            }
        }
    }
}