UVa1628 UVaLive5847 Pizza Delivery

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？

 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<iostream>

 using namespace std;

 void setIO(const string& s) {
     freopen((s + ".in").c_str(), "r", stdin);
     freopen((s + ".out").c_str(), "w", stdout);
 }
 template<typename Q> Q read(Q& x) {
     static char c, f;
     for(f = ; c = getchar(), !isdigit(c); ) if(c == '-') f = ;
     for(x = ; isdigit(c); c = getchar()) x = x *  + c - '';
     if(f) x = -x;
     return x;
 }
 template<typename Q> Q read() {
     static Q x; read(x); return x;
 }

 const int N =  + , INF = 0x3f3f3f3f;

 int p[N], v[N], n, cas;
 int f[N][N][N][];
 int vis[N][N][N][];

 int dfs(int s, int e, int cnt, int pos) {
     if(!cnt) return ;
     int &ans = f[s][e][cnt][pos];
     if(vis[s][e][cnt][pos] == cas) return ans;
     vis[s][e][cnt][pos] = cas;

     ans = -INF;
     int a[] = {s, e};

     /*for(int i = 0; i < s; i++) {
         ans = max(ans, v[i] - cnt * abs(p[i] - p[a[pos]]) + dfs(i, e, cnt - 1, 0));
     }
     for(int i = e + 1; i < n; i++) {
         ans = max(ans, v[i] - cnt * abs(p[i] - p[a[pos]]) + dfs(s, i, cnt - 1, 1));
     }*/

     if(s -  >= ) {
         int dlt = cnt * abs(p[s - ] - p[a[pos]]);
         ans = max(ans, v[s - ] - dlt + dfs(s - , e, cnt - , ));
         ans = max(ans, -dlt + dfs(s - , e, cnt, ));
     }
     if(e +  < n) {
         int dlt = cnt * abs(p[e + ] - p[a[pos]]);
         ans = max(ans, v[e + ] - dlt + dfs(s, e + , cnt - , ));
         ans = max(ans, -dlt + dfs(s, e + , cnt, )) ;
     }

     return ans;
 }

 int main() {
 #ifdef DEBUG
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w", stdout);
 #endif

     int T; scanf("%d", &T);
     for(cas = ; cas <= T; cas++) {
         scanf("%d", &n);
         for(int i = ; i < n; i++) scanf("%d", p + i);
         for(int i = ; i < n; i++) scanf("%d", v + i);

         int ans = ;
         for(int k = ; k <= n; k++) {
             for(int i = ; i < n; i++) {
                 ans = max(ans, v[i] - k * abs(p[i]) + dfs(i, i, k - , ));
             }
         }
         printf("%d\n", ans);
     }

     return ;
 }