cf237，D.Minesweeper 1D （小范围后效性dp）

题目：一维扫雷的情景，输入一行串，包含0,1,2，*，？

0，1， 2,表示周围的雷的个数，*表示雷，？表示未定。问共有多少中合理组合.

线性dp：

dp[n][i]表示的是长度为n的串的状态

i= 0时表示，其后面不加*的组合数；i= 1时，其后面加*的组合数；i= 2表示，当前的位置为*的组合数

注意初始化和最后的取解

Codefoeces 404D Minesweeper 1D "小范围后效性"dp写的不错

//#pragma warning (disable: 4786)//#pragma comment (linker, "/STACK:16777216")//HEAD#include <cstdio>#include <ctime>#include <cstdlib>#include <cstring>#include <queue>#include <string>#include <set>#include <stack>#include <map>#include <cmath>#include <vector>#include <iostream>#include <algorithm>using namespace std;//LOOP#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define FD(i, b, a) for(int i = (b); i>= (a); --i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(A,value) memset(A,value,sizeof(A))#define CPY(a, b) memcpy(a, b, sizeof(a))#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)//INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RS(s) scanf("%s", s)//OUTPUT#define WI(n) printf("%d\n", n)#define WS(s) printf("%s\n", s)typedef long long LL;const int INF = 1000000007;const double eps = 1e-10;const int maxn = 1100010;const int MOD = 1000000007;int n, k;int dp[maxn][3];int main (){    char s[maxn];    scanf("%s", s);    n = strlen(s);    if (s[0] == '1') dp[1][1] = 1;    else if (s[0] == '0') dp[1][0] = 1;    else if (s[0] == '*') dp[1][2] = 1;    else if (s[0] == '?')    {        dp[1][1] = 1;        dp[1][0] = 1;        dp[1][2] = 1;    }    for (int i = 1; i < n; i++)    {        if (s[i] == '?' || s[i] == '0')        {            dp[i + 1][0] += dp[i][0];            dp[i + 1][0] %= MOD;        }        if (s[i] == '?' || s[i] == '1')        {            dp[i + 1][0] += dp[i][2];            dp[i + 1][0] %= MOD;            dp[i + 1][1] += dp[i][0];            dp[i + 1][1] %= MOD;        }        if (s[i] == '?' || s[i] == '2')        {            dp[i + 1][1] += dp[i][2];            dp[i + 1][1] %= MOD;        }        if (s[i] == '?' || s[i] == '*')        {            dp[i + 1][2] += dp[i][1] + dp[i][2];            dp[i + 1][2] %= MOD;        }    }    cout << (dp[n][0] + dp[n][2]) % MOD << endl;    return 0;}

小范围后效性dp，参考上文写的。感觉记录后面的状态和不记录后面的状态的写法差不多，关键还是分析清楚

//#pragma warning (disable: 4786)//#pragma comment (linker, "/STACK:16777216")//HEAD#include <cstdio>#include <ctime>#include <cstdlib>#include <cstring>#include <queue>#include <string>#include <set>#include <stack>#include <map>#include <cmath>#include <vector>#include <iostream>#include <algorithm>using namespace std;//LOOP#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define FD(i, b, a) for(int i = (b); i>= (a); --i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(A,value) memset(A,value,sizeof(A))#define CPY(a, b) memcpy(a, b, sizeof(a))#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)//INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RS(s) scanf("%s", s)//OUTPUT#define WI(n) printf("%d\n", n)#define WS(s) printf("%s\n", s)typedef long long LL;const int INF = 1000000007;const double eps = 1e-10;const int MAXN = 1000010;const int MOD = 1000000007;int dp[1000100][5];char s[1000100];/***0:01:当前为1，前一个为*2:当前为1，后一个为*（前一个不为*）3:24:**/int main (){    scanf("%s", s);    int n = strlen(s);    if (s[0] == '2')    {        printf("0\n");        return 0;    }    if (s[0] == '0')        dp[1][0] = 1;    else if (s[0] == '1')        dp[1][2] = 1;    else if (s[0] == '*')        dp[1][4] = 1;    else if (s[0] = '?')    {        dp[1][0] = dp[1][2] = dp[1][4] = 1;    }    for (int i = 1; i < n; i++)    {        char c = s[i];        if (c == '1')        {            dp[i + 1][2] += (dp[i][0] + dp[i][1]) % MOD;            dp[i + 1][2] %= MOD;            dp[i + 1][1] += dp[i][4];            dp[i + 1][1] %= MOD;        }        else if (c == '0')        {            dp[i + 1][0] += (dp[i][0] + dp[i][1]) % MOD;            dp[i + 1][0] %= MOD;        }        else if (c == '2')        {            dp[i + 1][3] += dp[i][4];            dp[i + 1][3] %= MOD;        }        else if (c == '*')        {            dp[i + 1][4] += ((dp[i][3] + dp[i][2]) % MOD + dp[i][4]) % MOD;            dp[i + 1][4] %= MOD;        }        else        {            dp[i + 1][2] += (dp[i][0] + dp[i][1]) % MOD;            dp[i + 1][2] %= MOD;            dp[i + 1][1] += dp[i][4];            dp[i + 1][1] %= MOD;            dp[i + 1][0] += (dp[i][0] + dp[i][1]) % MOD;            dp[i + 1][0] %= MOD;            dp[i + 1][3] += dp[i][4];            dp[i + 1][3] %= MOD;            dp[i + 1][4] += ((dp[i][3] + dp[i][2]) % MOD + dp[i][4]) % MOD;            dp[i + 1][4] %= MOD;        }    }    int ans = dp[n][0];    ans += (dp[n][1] + dp[n][4]) % MOD;    ans %= MOD;    cout << ans << endl;    return 0;}