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：
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<cmath> 5 #define B (int(sqrt(n))) 6 #define N 200000 7 using namespace std; 8 struct nod{ 9 int num; 10 int next,cost; 11 }a[N+5]; 12 int n,m; 13 int main(void){ 14 scanf("%d",&n); 15 for(int i=0;i<n;++i) 16 scanf("%d",&a[i].num); 17 for(int i=n-1;i>=0;--i){ 18 int index=i+a[i].num; 19 int R=min(n,((i/B)+1)*B); 20 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**/ 21 a[i].cost=1; 22 a[i].next=index; 23 }else{ 24 a[i].cost=a[index].cost+1; 25 a[i].next=a[index].next; 26 } 27 } 28 scanf("%d",&m); 29 while(m--){ 30 int type; 31 scanf("%d",&type); 32 if(type==1){ 33 int index,sum=0; 34 scanf("%d",&index); 35 while(index<n){ 36 sum+=a[index].cost; 37 index=a[index].next; 38 } 39 printf("%d\n",sum); 40 }else if(type==2){ 41 int index,num; 42 scanf("%d%d",&index,&num); 43 int block=index/B; 44 a[index].num=num; 45 for(int i=index;i>=block*B;--i){ 46 int t=a[i].num+i; 47 int R=min(n,(block+1)*B); 48 if(t>=R){ 49 a[i].cost=1; 50 a[i].next=t; 51 }else{ 52 a[i].cost=a[t].cost+1; 53 a[i].next=a[t].next; 54 } 55 } 56 } 57 } 58 }
