BZOJ2958 序列染色

果然清华集训的题目。。。显然的DP题但是不会做。。。

我们令f[i][j][w]表示状态方程

w表示到了字符串的第w个

i = 0, 1, 2分别表示k个B和k个W都没填上、k个B填上了k个W没填上、k个B和k个W都填上了三种状态

j = 0, 1分别表示第w位上填B/W

于是方程就比较容易列出来了，注意要用到容斥原理

 /**************************************************************
     Problem: 2958
     User: rausen
     Language: C++
     Result: Accepted
     Time:652 ms
     Memory:33032 kb
 ****************************************************************/

 #include <cstdio>

 using namespace std;
 const int N = ;
 const int mod = 1e9 + ;

 int n, k;
 char st[N];
 int s0[N], s1[N], f[][][N];

 void read_in() {
   int i;
   char ch = getchar();
   while (ch != 'W' && ch != 'B' && ch != 'X')
     ch = getchar();
   for (i = ; i <= n; ++i)
     st[i] = ch, ch = getchar();
 }

 inline int calc(char c, int *s, int i, int t) {
   if (i < k || st[i - k] == c || s[i] - s[i - k]) return ;
   return f[t - ][i == k ?  :  - t][i - k];
 }

 int main() {
   int i;
   scanf("%d%d", &n, &k);
   read_in();
   f[][][] = ;
   for (i = ; i <= n; ++i) {
     s0[i] = s0[i - ] + (st[i] == 'W');
     s1[i] = s1[i - ] + (st[i] == 'B');
     if (st[i] != 'W') {
       f[][][i] = (0ll + f[][][i - ] + f[][][i - ] - calc('B', s0, i, ) + mod) % mod;
       f[][][i] = (0ll + f[][][i - ] + f[][][i - ] + calc('B', s0, i, )) % mod;
       f[][][i] = (0ll + f[][][i - ] + f[][][i - ]) % mod;
     }
     if (st[i] != 'B') {
       f[][][i] = (0ll + f[][][i - ] + f[][][i - ]) % mod;
       f[][][i] = (0ll + f[][][i - ] + f[][][i - ] - calc('W', s1, i, ) + mod) % mod;
       f[][][i] = (0ll + f[][][i - ] + f[][][i - ] + calc('W', s1, i, )) % mod;
     }
   }
   printf("%d\n", (f[][][n] + f[][][n]) % mod);
   return ;
 }

(p.s. Orz 江哥...)