NBOJv2 1050 Just Go（线段树/树状数组区间更新单点查询）

Problem 1050: Just Go

Time Limits:  3000 MS   Memory Limits:  65536 KB

64-bit interger IO format:  %lld   Java class name:  Main

Description

There is a river, which contains n stones from left to right. These stones are magic, each

one has a magic number Ai which means if you stand on the ith stone, you can jump to (i +1)th stone, (i+2)th stone, ..., (i+Ai)th stone(when i+Ai > n, you can only reach as far as n), at first, you stand on 1th stone, you want to calculate the number of ways to reach the nth stone.

Notice: you can not jump from right to left! 

Input

Input starts with an integer T(1 <= T <= 10), denoting the number of test cases. Each test case contains an integer n(1 <= n <= 1e5), denoting the number stones. Next line contains n integers Ai(1 <= Ai <= 1e8). 

Output

For each test case, print the number of way to reach the nth stone module 1e9+7. 

Sample Input

351 2 3 4 511022 1

Output for Sample Input

311

校赛那会儿的题目，主要操作就是区间更新、单点查询，树状数组和线段树都可以，树状数组的简单很多也好写很多，线段树嘛，线段树的模版题自行体会。

主要解题思路：刚开始肯定是1号石头初始化为1，其他的都是0，这个相信很好理解，然后就是往后覆盖区间，但却不是简单的覆盖，为了弄懂到底如何操作举个例子先，比如我到3号有a种路线，到4号有几种（假设4号只与3号连通）？当然跟3号一样，就是a种，如果3号可以跳跃的步数不止1即可以跳到5号上呢？此时会变成P5=P4+P3，即3号过来有三种，4号过来有1种（只能跳一步过来）。加起来就是4种，3->5一种，3->4>-5三种，推广出去就是不停地往后覆盖自己这个点的可到达情况数就好了，最后的答案当然就是Pn

树状数组代码：

#include<iostream>#include<algorithm>#include<cstdlib>#include<sstream>#include<cstring>#include<cstdio>#include<string>#include<deque>#include<stack>#include<cmath>#include<queue>#include<set>#include<map>using namespace std;#define INF 0x3f3f3f3f#define MM(x,y) memset(x,y,sizeof(x))#define LC(x) (x<<1)#define RC(x) ((x<<1)1)#define MID(x,y) ((x+y)>>1)typedef pair<int,int> pii;typedef long long LL;const double PI=acos(-1.0);const int N=100010;const LL mod=1000000007;LL tree[N],arr[N];inline int lowbit(int k){return k&(-k);}void add(int k,LL val){while (k<=100000){tree[k]+=val;k+=lowbit(k);}}LL getsum(int k){LL r=0;while (k){r+=tree[k];r%=mod;k-=lowbit(k);}return r%mod;}void init(){MM(tree,0);MM(arr,0);}int main(void){int tcase,i,j,l,r,n;scanf("%d",&tcase);while (tcase--){scanf("%d",&n);init();add(1,1);add(2,-1);for (i=1; i<=n; i++){scanf("%I64d",&arr[i]);}for (i=1; i<=n; i++){ l=i+1;r=i+arr[i];if(r>n)r=n;LL pres=getsum(i);add(l,pres);add(r+1,-pres);}printf("%I64d\n",getsum(n));}return 0;}

线段树代码：

#include<iostream>#include<algorithm>#include<cstdlib>#include<sstream>#include<cstring>#include<cstdio>#include<string>#include<deque>#include<stack>#include<cmath>#include<queue>#include<set>#include<map>using namespace std;#define INF 0x3f3f3f3f#define MM(x,y) memset(x,y,sizeof(x))#define LC(x) (x<<1)#define RC(x) ((x<<1)+1)#define MID(x,y) ((x+y)>>1)typedef pair<int,int> pii;typedef long long LL;const double PI=acos(-1.0);const int N=100010;const LL mod=1e9+7;struct info{LL l,mid,r;LL sum,add;};info T[N<<2];LL arr[N];void pushup(int k){T[k].sum=T[LC(k)].sum+T[RC(k)].sum;}void pushdown(int k){T[RC(k)].add+=T[k].add;T[RC(k)].sum+=T[k].add*(T[RC(k)].r-T[RC(k)].l+1);T[LC(k)].add+=T[k].add;T[LC(k)].sum+=T[k].add*(T[LC(k)].r-T[LC(k)].l+1);T[k].add=0;}void build(int k,LL l,LL r){T[k].l=l;T[k].r=r;T[k].mid=MID(T[k].l,T[k].r);T[k].add=0;T[k].sum=0;if(l==r)T[k].sum=(l==1&&r==1?1:0);else{build(LC(k),l,T[k].mid);build(RC(k),T[k].mid+1,r);pushup(k);}}void update(int k,LL l,LL r,LL val){if(r<T[k].l||l>T[k].r)return ;if(l<=T[k].l&&r>=T[k].r){T[k].add+=val;T[k].sum+=val*(T[k].r-T[k].l+1);T[k].sum%=mod;}else{if(T[k].add)pushdown(k);update(LC(k),l,r,val);update(RC(k),l,r,val);pushup(k);}}LL query(int k,LL x){if(T[k].l==T[k].r&&T[k].l==x)return T[k].sum%mod;if(T[k].add)pushdown(k);if(x<=T[k].mid)return query(LC(k),x)%mod;else if(x>T[k].mid)return query(RC(k),x)%mod;}int main(void){int tcase,i,j,n;LL l,r,x,val;scanf("%d",&tcase);while (tcase--){scanf("%d",&n);MM(arr,0);for (i=1; i<=n; i++)scanf("%I64d",&arr[i]);build(1,1,n);for (i=1; i<=n; i++)update(1,(LL)(i+1),(LL)(i+arr[i]>n?n:i+arr[i]),query(1,i));printf("%I64d\n",query(1,n)%mod);}return 0;}