UVALive 7276 Wooden Signs （DP）

Wooden Signs

题目链接：

http://acm.hust.edu.cn/vjudge/contest/127406#problem/E

Description

http://7xjob4.com1.z0.glb.clouddn.com/0f10204481da21e62f8c145939e5828e

Input

The input file contains several test cases, each of them as described below.
The first line has one integer N and the second line contains a permutation of the integers from 1
to N + 1. Integers in the same line are separated by a single space.
Constraints:
1 ≤ N

Output

For each test case, the output has a single line with the number (modulo 2
31 − 1 = 2147483647) of
distinct signs that can be described by the given permutation.

Sample Input

5
2 6 5 1 4 3
2
2 3 1
20
3 21 10 15 6 9 2 5 20 13 17 19 14 7 11 18 16 12 1 8 4

Sample Output

6
1
1887


##题意：

木匠要做N个路标并把它们钉在一起(如图).
每个路标可以朝左也可以朝右，但是每个路标的起点要跟下一层路标的起点或终点重合. 重合位置必须有面积覆盖，不能像231右图那种只重合一个点.
现在他不记得这些路标各自的朝向了. 但是他记得N+1个值，其中第i个值是第i个路标的起点坐标. (第N+1为最后一块的终点).
求可能满足条件的路标有多少种情况. (区别在于各自的朝向)(第一块路标一定朝右)


##题解：

考虑每块路标的终点：
若第i块路标的坐标区间为[L, R], 第i+1块路标的终点为Xi+1.
那么基于第i块路标来摆放第i+1块的方式有两种：
①第i+1块的起点在L处(朝右). 这要求 Xi+1 > L .
②第i+1块的起点在R处(朝左). 这要求 Xi+1

##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 2016
#define mod 2147483647LL
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;

LL dp[maxn][2][maxn];

int n, arrow[maxn];

int main(int argc, char const *argv[])

{

//IN;

while(scanf("%d", &n) != EOF)
{
    int leftmost; scanf("%d", &leftmost);
    for(int i=1; i<=n; i++)
        scanf("%d", &arrow[i]);
    memset(dp, 0, sizeof(dp));
    dp[1][0][leftmost] = 1;
    for(int i=1; i<n; i++) {
        for(int j=1; j<=n+1; j++) {
            if(dp[i][0][j]) {
                if(arrow[i+1] > j) dp[i+1][0][j] = (dp[i+1][0][j] + dp[i][0][j]) % mod;
                if(arrow[i+1] < arrow[i]) dp[i+1][1][arrow[i]] = (dp[i+1][1][arrow[i]] + dp[i][0][j]) % mod;
            }
            if(dp[i][1][j]) {
                if(arrow[i+1] < j) dp[i+1][1][j] = (dp[i+1][1][j] + dp[i][1][j]) % mod;
                if(arrow[i+1] > arrow[i]) dp[i+1][0][arrow[i]] = (dp[i+1][0][arrow[i]] + dp[i][1][j]) % mod;
            }
        }
    }
    LL ans = 0;
    for(int j=1; j<=n+1; j++) {
        ans = (ans + dp[n][0][j]) % mod;
        ans = (ans + dp[n][1][j]) % mod;
    }
    printf("%lld\n", ans);
}
return 0;

}