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+];
 int n,m;
 int main(void){
     scanf("%d",&n);
     for(int i=;i<n;++i)
         scanf("%d",&a[i].num);
     for(int i=n-;i>=;--i){
         int index=i+a[i].num;
         int R=min(n,((i/B)+)*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=;
             a[i].next=index;
         }else{
             a[i].cost=a[index].cost+;
             a[i].next=a[index].next;
         }
     }
     scanf("%d",&m);
     while(m--){
         int type;
         scanf("%d",&type);
         if(type==){
             int index,sum=;
             scanf("%d",&index);
             while(index<n){
                 sum+=a[index].cost;
                 index=a[index].next;
             }
             printf("%d\n",sum);
         }else if(type==){
             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+)*B);
                 if(t>=R){
                     a[i].cost=;
                     a[i].next=t;
                 }else{
                     a[i].cost=a[t].cost+;
                     a[i].next=a[t].next;
                 }
             }
         }
     }
 }