Codeforces 479E. Riding in a Lift DP
来源:互联网 发布:携程首席数据官 编辑:程序博客网 时间:2024/06/05 17:54
O(n^3) 的dp , 用前缀和减少到O(n^2)
Imagine that you are in a building that has exactly n floors. You can move between the floors in a lift. Let's number the floors from bottom to top with integers from 1 to n. Now you're on the floor number a. You are very bored, so you want to take the lift. Floor number b has a secret lab, the entry is forbidden. However, you already are in the mood and decide to make k consecutive trips in the lift.
Let us suppose that at the moment you are on the floor number x (initially, you were on floor a). For another trip between floors you choose some floor with number y (y ≠ x) and the lift travels to this floor. As you cannot visit floor b with the secret lab, you decided that the distance from the current floor x to the chosen y must be strictly less than the distance from the current floor x to floor b with the secret lab. Formally, it means that the following inequation must fulfill: |x - y| < |x - b|. After the lift successfully transports you to floor y, you write down number y in your notepad.
Your task is to find the number of distinct number sequences that you could have written in the notebook as the result of k trips in the lift. As the sought number of trips can be rather large, find the remainder after dividing the number by 1000000007 (109 + 7).
The first line of the input contains four space-separated integers n, a, b, k (2 ≤ n ≤ 5000, 1 ≤ k ≤ 5000, 1 ≤ a, b ≤ n, a ≠ b).
Print a single integer — the remainder after dividing the sought number of sequences by 1000000007 (109 + 7).
5 2 4 1
2
5 2 4 2
2
5 3 4 1
0
Two sequences p1, p2, ..., pk and q1, q2, ..., qk are distinct, if there is such integer j (1 ≤ j ≤ k), that pj ≠ qj.
Notes to the samples:
- In the first sample after the first trip you are either on floor 1, or on floor 3, because |1 - 2| < |2 - 4| and |3 - 2| < |2 - 4|.
- In the second sample there are two possible sequences: (1, 2); (1, 3). You cannot choose floor 3 for the first trip because in this case no floor can be the floor for the second trip.
- In the third sample there are no sought sequences, because you cannot choose the floor for the first trip.
/* ***********************************************Author :CKbossCreated Time :2015年03月07日 星期六 19时15分20秒File Name :CF479E.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;typedef long long int LL;const LL mod = 1e9+7LL;const int maxn=5050;LL n,a,b,k;LL dp[maxn][maxn];LL presum[maxn];int main(){ //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout);cin>>n>>a>>b>>k;for(int i=1;i<=n;i++) dp[k][i]=1;dp[k][b]=0;for(int i=1;i<=n;i++) presum[i]=(presum[i-1]+dp[k][i])%mod;for(int i=k-1;i>=0;i--){for(int j=1;j<=n;j++){if(j==b) continue;int L,R,dur=abs(j-b);if(j>b) { L=b+1; R=min((int)n,j+dur-1); }else if(j<b) { L=max(1,j-dur+1); R=b-1; }LL ret=((presum[R]-presum[L-1]+mod)%mod-dp[i+1][j]+mod)%mod;dp[i][j]=ret;}for(int j=1;j<=n;j++) presum[j]=(presum[j-1]+dp[i][j])%mod;}cout<<dp[0][a]<<endl; return 0;}
- codeforces 479e Riding in a Lift | dp
- [笔记] Codeforces#274 Riding in a Lift (479E) DP
- Codeforces 479E Riding in a Lift(dp)
- Codeforces 479E. Riding in a Lift DP
- Codeforces 479E Riding in a Lift (DP)
- Codeforces 479E Riding in a Lift DP+前缀和
- codeforces 479E Riding in a Lift dp+前缀数组
- CodeForces 479 E. Riding in a Lift
- Codeforces 479E Riding in a Lift
- CodeForces 479E E. Riding in a Lift
- codeforces 479E E. Riding in a Lift(dp+段修改的优化)
- codeforces 479E Riding in a Lift dp+前缀数组优化
- Codeforces 479E Riding in a Lift【Dp+前缀和优化+二分】好题~
- CF 479E Riding in a Lift 前缀和 DP
- Codeforce 题目479E Riding in a Lift(DP)
- codeforce 479E - Riding in a Lift
- codeforce 479E Riding in a Lift
- Codeforces Round #274 (Div. 2) E题:Riding in a Lift(DP)
- March我来了!
- 计算txt文本中单词的个数
- Unity3D游戏开发之 (Navmesh) 和寻路 (Pathfinding)烘焙
- 第5章3节《MonkeyRunner源码剖析》Monkey原理分析-启动运行: 启动脚本(原创)
- 堆排序示例:对n个整数进行排序.
- Codeforces 479E. Riding in a Lift DP
- Android学习总结七:绑定ListView数据的方式
- 蓝天碧水中国梦-中国制造(1)-2014董明珠代理商内部讲话
- hdu 2191 多重背包,自己写模板
- 蓝桥杯c/c++预选赛 地宫取宝
- 城市行--求网格中的可达路径数
- lambda expression demo
- Java中的自增操作符与中间缓存变量机制
- malloc和new的区别