有一个长度为 n 的 01 串,你可以每次将相邻的 k 个字符合并,得到一个新的字符并获得一定分数。得到的新字
符和分数由这 k 个字符确定。你需要求出你能获得的最大分数。
输出一个整数表示答案
考虑区间DP。记f[i][j][k]表示把[i,j]区间内的数都合并成k的最大价值。因为k<=8,所以可以用一个二进制数来把状态压起来。
考虑如何转移,注意到只有长度是k的线段才可以合并成一个,所以,长度在模k-1意义下余1的线段一定只会被合并成1个数,所以,我们每次转移的时候只要枚举长度在模k-1意义下为1的前(后)缀,然后再枚举2^k的状态即可。时间复杂度O(n^3*2^k/k^2)。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <cmath> 5 6 #ifdef WIN32 7 #define LL "%I64d" 8 #else 9 #define LL "%lld" 10 #endif 11 12 #ifdef CT 13 #define debug(...) printf(__VA_ARGS__) 14 #define setfile() 15 #else 16 #define debug(...) 17 #define filename "" 18 #define setfile() freopen(filename".in", "r", stdin); freopen(filename".out", "w", stdout) 19 #endif 20 21 #define R register 22 #define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 15, stdin), S == T) ? EOF : *S++) 23 #define dmax(_a, _b) ((_a) > (_b) ? (_a) : (_b)) 24 #define dmin(_a, _b) ((_a) < (_b) ? (_a) : (_b)) 25 #define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0) 26 #define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0) 27 #define cabs(_x) ((_x) < 0 ? (- (_x)) : (_x)) 28 char B[1 << 15], *S = B, *T = B; 29 inline int F() 30 { 31 R char ch; R int cnt = 0; R bool minus = 0; 32 while (ch = getc(), (ch < ‘0‘ || ch > ‘9‘) && ch != ‘-‘) ; 33 ch == ‘-‘ ? minus = 1 : cnt = ch - ‘0‘; 34 while (ch = getc(), ch >= ‘0‘ && ch <= ‘9‘) cnt = cnt * 10 + ch - ‘0‘; 35 return minus ? -cnt : cnt; 36 } 37 #define maxn 666 38 int str[maxn], to[maxn]; 39 long long f[310][310][256], val[maxn]; 40 int main() 41 { 42 // setfile(); 43 R int n = F(), k = F(); 44 for (R int i = 1; i <= n; ++i) 45 { 46 R char ch = getc(); 47 while (ch < ‘0‘ || ch > ‘1‘) ch = getc(); 48 str[i] = ch - ‘0‘; 49 } 50 for (R int i = 0; i < 1 << k; ++i) 51 { 52 to[i] = F(); 53 val[i] = F(); 54 } 55 memset(f, -63, sizeof (f)); R long long inf = f[0][0][0]; 56 for (R int i = 1; i <= n; ++i) f[i][i][str[i]] = 0; 57 for (R int len = 2; len <= n; ++len) 58 { 59 R int l = len - 1; while (l >= k) l -= k - 1; 60 for (R int i = 1, j = len; j <= n; ++i, ++j) 61 { 62 R long long *tmp = f[i][j]; 63 for (R int mid = j; mid > i; mid -= k - 1) 64 for (R int s = 0; s < (1 << l); ++s) 65 { 66 cmax(tmp[s << 1], f[i][mid - 1][s] + f[mid][j][0]); 67 cmax(tmp[s << 1 | 1], f[i][mid - 1][s] + f[mid][j][1]); 68 } 69 if (l == k - 1) 70 { 71 R long long g[2]; memset(g, -63, sizeof (g)); 72 for (R int s = 0; s < (1 << k); ++s) 73 cmax(g[to[s]], tmp[s] + val[s]); 74 tmp[0] = g[0]; tmp[1] = g[1]; 75 } 76 // if (i == 1 && j == 2) printf("%lld\n", f[i][j][0] ); 77 } 78 } 79 /* for (R int i = 1; i <= n; ++i) 80 for (R int j = i; j <= n; ++j) 81 for (R int s = 0; s < 1 << k; ++s) 82 printf("f[%d][%d][%d] = %lld\n", i, j, s, f[i][j][s] );*/ 83 R long long ans = 0; 84 for (R int i = 0; i < (1 << k); ++i) 85 cmax(ans, f[1][n][i]); 86 printf("%lld\n", ans ); 87 return 0; 88 } 89 /* 90 3 2 91 101 92 1 10 93 1 10 94 0 20 95 1 30 96 */
原文:http://www.cnblogs.com/cocottt/p/6685127.html