洛谷P4401 [IOI2007]Miners 矿工配餐
题意
题解
但是我太菜了，所以我是$\mathcal O({16}^2n)$ 的。
调试记录
- 没用滚动数组
- for (int l=0; l<16; l++) 打成了 for (int l=0; j<16; j++) ...
代码
1 #include <bits/stdc++.h> 2 #pragma GCC optimize(3) 3 #define up(x, y) x = max(x, y) 4 using namespace std; 5 6 char a[100005]; 7 int dp[2][16][16]; 8 9 inline int typ(char x) { 10 return x == 'M' ? 0 : (x == 'F' ? 1 : (x == 'B' ? 2 : 3)); 11 } 12 inline int sta(int pr, int pe) { 13 return pr * 4 + pe; 14 } 15 inline int E(int sta) { 16 return sta % 4; 17 } 18 inline int R(int sta) { 19 return sta / 4; 20 } 21 int kinds[3]; 22 inline int cnt(int sta, int x) { 23 int r = R(sta), e = E(sta); 24 kinds[0] = x; int p=0; 25 if (r != 3) kinds[++p] = r; 26 if (e != 3) kinds[++p] = e; 27 sort(kinds, kinds+p+1); 28 int m=1, z=kinds[0]; 29 for (int i=1; i<=p; i++) 30 if (kinds[i] != z) ++m, z = kinds[i]; 31 return m; 32 } 33 34 int main() { 35 int n; scanf("%d", &n); 36 scanf("%s", a+1); 37 memset(dp[0], 200, sizeof(dp[0])); 38 int cur = 0, nxt = 1; 39 dp[cur][sta(3, 3)][sta(3, 3)] = 0; 40 for (int i=1; i<=n; i++) { 41 memset(dp[nxt], 200, sizeof(dp[nxt])); 42 for (int j=0; j<16; j++) 43 for (int l=0; l<16; l++) { 44 if (R(j) != 3 && E(j) == 3) continue; 45 if (R(l) != 3 && E(l) == 3) continue; 46 int x = typ(a[i]); 47 up(dp[nxt][sta(E(j), x)][l], dp[cur][j][l] + cnt(j, x)); 48 up(dp[nxt][j][sta(E(l), x)], dp[cur][j][l] + cnt(l, x)); 49 } 50 cur = !cur; nxt = !nxt; 51 } 52 int ans = 0; 53 for (int i=0; i<16; i++) for(int j=0; j<16; j++) { 54 up(ans, dp[cur][i][j]); 55 } 56 printf("%d\n", ans); 57 return 0; 58 }
