HDU 4055 Number String dp

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4055

Number String

Time Limit: 10000/5000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others) #### 问题描述 > The signature of a permutation is a string that is computed as follows: for each pair of consecutive elements of the permutation, write down the letter 'I' (increasing) if the second element is greater than the first one, otherwise write down the letter 'D' (decreasing). For example, the signature of the permutation {3,1,2,7,4,6,5} is "DIIDID". > > Your task is as follows: You are given a string describing the signature of many possible permutations, find out how many permutations satisfy this signature. > > Note: For any positive integer n, a permutation of n elements is a sequence of length n that contains each of the integers 1 through n exactly once.

输入

Each test case consists of a string of 1 to 1000 characters long, containing only the letters 'I', 'D' or '?', representing a permutation signature.

Each test case occupies exactly one single line, without leading or trailing spaces.

Proceed to the end of file. The '?' in these strings can be either 'I' or 'D'.

输出

For each test case, print the number of permutations satisfying the signature on a single line. In case the result is too large, print the remainder modulo 1000000007.

样例输入

II
ID
DI
DD
?D
??

样例输出

1
2
2
1
3
6

题意

给你一个排列的相邻位置的大小关系：'I'表示前一个小于后一个，'D'表示前一个大于后一个，'?'表示不确定。问满足这些大小关系的全排列的所有可能数。

题解

比较容易想到的是dp[i][j]表示前i个数，最后一个是j的满足条件的总数，然后用前缀和优化下跑n^2，但是，这样跑出来的结果是可重集的答案！不是1到n的全排列！！！
那么如何做到每个数子都只出现一次？一种巧妙的做法是dp[i][j]表示1到i的全排列的符合条件的总数，那么怎么转移？其实转移没差，dp[i][j]=sigma(dp[i][k]),1<=k<=i,你可以这样理解：吧前面所有>=j的数都全部+1。比如{1,3,2，4}，现在插入一个3变成{1,4,2,5,3}！

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long  LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1111;
const int mod=1e9+7;
char str[maxn];

LL dp[maxn][maxn],sum[maxn][maxn];

int main() {
    while(scf("%s",str+2)==1) {
        clr(sum,0);
        sum[1][1]=dp[1][1]=1;
        int len=strlen(str+2);
//        bug(len);
        for(int i=2; i<len+2; i++) {
            for(int j=1; j<=i; j++) {
                if(str[i]=='I') {
                    dp[i][j]=sum[i-1][j-1];
                } else if(str[i]=='D') {
                    dp[i][j]=(sum[i-1][i-1]-sum[i-1][j-1]+mod)%mod;
                } else {
                    dp[i][j]=sum[i-1][i-1];
                }
                sum[i][j]=sum[i][j-1]+dp[i][j];
                sum[i][j]%=mod;
            }
        }
        prf("%lld
",sum[len+1][len+1]);
    }
    return 0;
}

//end-----------------------------------------------------------------------