http://codeforces.com/contest/404/problem/D
Game "Minesweeper 1D" is played on a line of squares, the line's height is 1 square, the line's width is n squares. Some of the squares contain bombs. If a square doesn't contain a bomb, then it contains a number from 0 to 2 — the total number of bombs in adjacent squares.
For example, the correct field to play looks like that: 001*2***101*. The cells that are marked with "*" contain bombs. Note that on the correct field the numbers represent the number of bombs in adjacent cells. For example, field 2* is not correct, because cell with value 2 must have two adjacent cells with bombs.
Valera wants to make a correct field to play "Minesweeper 1D". He has already painted a squared field with width of n cells, put several bombs on the field and wrote numbers into some cells. Now he wonders how many ways to fill the remaining cells with bombs and numbers are there if we should get a correct field in the end.
Input
The first line contains sequence of characters without spaces s1s2... sn (1 ≤ n ≤ 106), containing only characters "*", "?" and digits "0", "1" or "2". If character si equals "*", then the i-th cell of the field contains a bomb. If character si equals "?", then Valera hasn't yet decided what to put in the i-th cell. Character si, that is equal to a digit, represents the digit written in the i-th square.
Output
Print a single integer — the number of ways Valera can fill the empty cells and get a correct field.
As the answer can be rather large, print it modulo 1000000007 (109 + 7).
題意：長度爲n的序列，每個元素是*或者0/1/2或者？，*表示這裏埋了一顆炸彈，數字表示這個位置的左右兩邊位置有0/1/2個炸彈，？表示這裏實際是什麼不知道。給定一個序列，求合法的方案數。：：:
思路:
①自己想的：
設f(i,0/1/3):只考慮前i個位置，並且第i個位置是0/1/*的方案數(因爲不考慮i+1，所以i位置不可能是2)
然後用了很多的if...else...來判斷轉移，搞得邏輯實際上十分複雜，見代碼註釋吧。
#include<bits/stdc++.h>
using namespace std;
#define maxn 1000000+1000
#define mod 1000000007
#define ll long long
ll n,f[maxn][4];
char s[maxn];
void dp()
{
f[0][0]=1; //只考慮前0個位置，只有"什麼都沒有"是合理的情況
if(s[1]=='?')f[1][0]=f[1][3]=1; //後面要轉移到f(i-2)，所有i=1也要初始化
else if(s[1]=='*')f[1][3]=1;
else if(s[1]=='0')f[1][0]=1;
for(int i=2;i<=n;i++)
{
if(s[i]=='?')
{
f[i][0]=(f[i-1][0]+f[i-1][1])%mod; //(i是0)i-1,i : 0,0or1,0; 2,0and*,0不合法
ll last=0;//(i是*) //i-2,i-1,i: 0/1,1,* [注意這裏i-1是1,在f(i-1)時這裏是0，因爲不考慮i]
if(s[i-1]=='1'||s[i-1]=='?')last=f[i-2][0]+f[i-2][1];
// i-2,i-1,i: *,2,* //0/1,1,* //0/1/*,*,*
f[i][3]=((s[i-1]=='2'||s[i-1]=='?'?f[i-2][3]:0)+last+(s[i-1]=='*'||s[i-1]=='?'?f[i-1][3]:0))%mod;
f[i][1]=f[i-1][3]; //(i是1)i-1,i: *,1,其餘不合法
}
else if(s[i]=='*') //下面與上面是一樣的
{
ll last=0;
if(s[i-1]=='1'||s[i-1]=='?')last=f[i-2][0]+f[i-2][1];
f[i][3]=((s[i-1]=='2'||s[i-1]=='?'?f[i-2][3]:0)+last+(s[i-1]=='*'||s[i-1]=='?'?f[i-1][3]:0))%mod;
}
else if(s[i]=='0')f[i][0]=f[i-1][0]+f[i-1][1];
else if(s[i]=='1')f[i][1]=f[i-1][3];
}
}
int main()
{
//freopen("input.in","r",stdin);
scanf("%s",s+1);
n=strlen(s+1);
dp();
cout<<(f[n][0]+f[n][1]+f[n][2]+f[n][3])%mod<<endl;
return 0;
}
②官方題解：
設f(i,type):只考慮前i個位置，第i個位置type分別是：
0.填0
1.填1，*在左
2.填1，*在右
3.填2
4.填*
這樣子轉移就自然多了，邏輯很清晰(用到了一點未來計算的思想)
#include<bits/stdc++.h>
using namespace std;
#define maxn 1000000+1000
#define mod 1000000007
#define ll long long
ll n,f[maxn][5];
char s[maxn];
void dp()
{
f[0][0]=f[0][2]=1;
for(int i=1;i<=n;i++)
{
if(s[i]=='?')
{
f[i][0]=f[i-1][0]+f[i-1][1],f[i][0]%=mod;
f[i][1]=f[i-1][4];
f[i][2]=f[i-1][0]+f[i-1][1],f[i][2]%=mod;
f[i][3]=f[i-1][4];
f[i][4]=f[i-1][4]+f[i-1][2]+f[i-1][3],f[i][4]%=mod;
}
else if(s[i]=='0')f[i][0]=f[i-1][0]+f[i-1][1],f[i][0]%=mod;
else if(s[i]=='*')f[i][4]=f[i-1][4]+f[i-1][2]+f[i-1][3],f[i][4]%=mod;
else if(s[i]=='2')f[i][3]=f[i-1][4];
else if(s[i]=='1')
{
f[i][1]=f[i-1][4];
f[i][2]=f[i-1][0]+f[i-1][1],f[i][2]%=mod;
}
}
}
int main()
{
//freopen("input.in","r",stdin);
scanf("%s",s+1);
n=strlen(s+1);
dp();
cout<<(f[n][0]+f[n][1]+f[n][4])%mod<<endl;
return 0;
}