[笔记] Codeforces#274 Riding in a Lift (479E) DP

http://codeforces.com/problemset/problem/479/E

dp[i][j] 表示第i次到达j层的路径数量

转移方程 

dp[i][j] = sum(dp[i-1][k]) , (k是满足要求的起始层)

由于k的集合组成一个连续的区间(剔除k=j), 故可以预先计算出区间左右边界,并维护一个部分和O(1)求出

sum(dp[i-1][k]) = sum[i-1][right] - sum[i-1][left-1] - dp[i-1][j]

使用滚动数组

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
using namespace std;

const int MAX = 5005;
const int MOD = 1000000007;
int dp[2][MAX];
int sum[2][MAX];

int main()
{
int n,a,b,k;
cin>>n>>a>>b>>k;
memset(dp,0,sizeof(dp));
memset(sum,0,sizeof(sum));
int sw = 0;
dp[sw][a] = 1;
for (int i=1;i<=n;i++) sum[sw][i] = sum[sw][i-1]+dp[sw][i];

for (int i=1;i<=k;i++){
for (int x=1;x<=n;x++){
if (x==b) continue;
int left,right;
if (x<b){
left = 1;
right = (x+b)/2;
if ((x+b)%2==0) right--;
}else {
left = (x+b)/2+1;
right = n;
}

dp[sw^1][x] = ((sum[sw][right] - sum[sw][left-1] -dp[sw][x])%MOD+MOD)%MOD;
sum[sw^1][x] = (sum[sw^1][x-1] + dp[sw^1][x])%MOD;
}
sw^=1;
}

int ans = 0;
for (int i=1;i<=n;i++) ans = (ans+dp[sw][i])%MOD;
printf("%d\n",ans);

return 0;
}