填坑系列(p.302)
既然不知道后面还要卖多少个就加一维状态嘛。。
lrj写的O(n)转移？其实转移可以O(1)
貌似按x排序有奇效？
1 #include<cstdio> 2 #include<cstring> 3 #include<cstdlib> 4 #include<algorithm> 5 #include<iostream> 6 7 using namespace std; 8 9 void setIO(const string& s) { 10 freopen((s + ".in").c_str(), "r", stdin); 11 freopen((s + ".out").c_str(), "w", stdout); 12 } 13 template<typename Q> Q read(Q& x) { 14 static char c, f; 15 for(f = 0; c = getchar(), !isdigit(c); ) if(c == '-') f = 1; 16 for(x = 0; isdigit(c); c = getchar()) x = x * 10 + c - '0'; 17 if(f) x = -x; 18 return x; 19 } 20 template<typename Q> Q read() { 21 static Q x; read(x); return x; 22 } 23 24 const int N = 100 + 10, INF = 0x3f3f3f3f; 25 26 int p[N], v[N], n, cas; 27 int f[N][N][N][2]; 28 int vis[N][N][N][2]; 29 30 int dfs(int s, int e, int cnt, int pos) { 31 if(!cnt) return 0; 32 int &ans = f[s][e][cnt][pos]; 33 if(vis[s][e][cnt][pos] == cas) return ans; 34 vis[s][e][cnt][pos] = cas; 35 36 ans = -INF; 37 int a[] = {s, e}; 38 39 /*for(int i = 0; i < s; i++) { 40 ans = max(ans, v[i] - cnt * abs(p[i] - p[a[pos]]) + dfs(i, e, cnt - 1, 0)); 41 } 42 for(int i = e + 1; i < n; i++) { 43 ans = max(ans, v[i] - cnt * abs(p[i] - p[a[pos]]) + dfs(s, i, cnt - 1, 1)); 44 }*/ 45 46 if(s - 1 >= 0) { 47 int dlt = cnt * abs(p[s - 1] - p[a[pos]]); 48 ans = max(ans, v[s - 1] - dlt + dfs(s - 1, e, cnt - 1, 0)); 49 ans = max(ans, -dlt + dfs(s - 1, e, cnt, 0)); 50 } 51 if(e + 1 < n) { 52 int dlt = cnt * abs(p[e + 1] - p[a[pos]]); 53 ans = max(ans, v[e + 1] - dlt + dfs(s, e + 1, cnt - 1, 1)); 54 ans = max(ans, -dlt + dfs(s, e + 1, cnt, 1)) ; 55 } 56 57 return ans; 58 } 59 60 int main() { 61 #ifdef DEBUG 62 freopen("in.txt", "r", stdin); 63 freopen("out.txt", "w", stdout); 64 #endif 65 66 int T; scanf("%d", &T); 67 for(cas = 1; cas <= T; cas++) { 68 scanf("%d", &n); 69 for(int i = 0; i < n; i++) scanf("%d", p + i); 70 for(int i = 0; i < n; i++) scanf("%d", v + i); 71 72 int ans = 0; 73 for(int k = 1; k <= n; k++) { 74 for(int i = 0; i < n; i++) { 75 ans = max(ans, v[i] - k * abs(p[i]) + dfs(i, i, k - 1, 0)); 76 } 77 } 78 printf("%d ", ans); 79 } 80 81 return 0; 82 }