codeforces Round #237(div2) D解题报告

D. Minesweeper 1D

time limit per test
2 seconds

memory limit per test
512 megabytes

input
standard input

output
standard output

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).

Sample test(s)

input

?01???

output

4

input

?

output

2

input

**12

output

0

input

1

output

0

Note

In the first test sample you can get the following correct fields: 001**1, 001***, 001*2*, 001*10.

题目大意:

        扫雷游戏的一维版(也就是说在一条线上,不是常玩的矩阵类型)

        每个格子有几种情况:

                '?'  表示还没有填充

               '0'  表示左右两边都没有雷

               '1'  表示两边有某个地方有雷

               '2'  表示两边都有雷

               '*'  表示该地方就是一个雷

        求出有多少种不同的填充'?'的方案

解法:

       本题有递推关系,即i位置的放置情况,跟前一个雷有关.然而有部分不怎么符合严格递推,那就是'1'这个情况,有可能是左边有雷,也有可能是右边有雷.那我们就把'1'这种情况拆开来,即左边有雷和右边有雷.然而就可以写出递推关系式:

               0表示两边没有雷;

               1表示右边有雷;

               2表示左边有雷;

               3表示两边都有雷;

               4表示当前位置有雷;

               dp[i][0] = dp[i-1][0] + dp[i-1][2];

               dp[i][1] = dp[i-1][0] + dp[i-1][2];

               dp[i][2] = dp[i-1][4];

               dp[i][3] = dp[i-1][4];

               dp[i][4] = dp[i-1][3] + dp[i-1][4] + dp[i-1][1];

               递推前置条件: dp[0][0] = 1, dp[0][2] = 1;

       最后输出最有一个格子的所有总方案数,由于是最后一个,所以不存在1右边有雷和3两边有雷的情况.

做题的时候,由于判断情况较多(多了一个'?'),写的很乱,网上看到某位写的代码比较简练而且巧妙,所以贴他的代码好了

#include <cstdio>
#include <cstring>

using namespace std;

const int INF = 1e9+7;

int n;
int dp[1000010][5];
char st[1000010];

void init() {
scanf("%s",st+1);
n = strlen(st+1);
}

void add(int &x, int y) {
x += y;
if (x >= INF) x -= INF;
}

void solve() {
dp[0][0] = dp[0][1] = 1;

for (int i = 1; i <= n; i++) {
bool g = (st[i] == '?');

if (g || st[i] == '0') {
add(dp[i][0], dp[i-1][0]);
add(dp[i][0], dp[i-1][2]);
}
if (g || st[i] == '1') {
add(dp[i][1], dp[i-1][0]);
add(dp[i][1], dp[i-1][2]);
add(dp[i][2], dp[i-1][4]);
}
if (g || st[i] == '2') {
add(dp[i][3], dp[i-1][4]);
}
if (g || st[i] == '*') {
add(dp[i][4], dp[i-1][4]);
add(dp[i][4], dp[i-1][1]);
add(dp[i][4], dp[i-1][3]);
}
}

int ans = 0;
bool g = (st
== '?');
if (g || st
== '0') add(ans, dp
[0]);
if (g || st
== '1') add(ans, dp
[2]);
if (g || st
== '*') add(ans, dp
[4]);
printf("%d\n", ans);
}

int main() {
init();
solve();
}