1628 Pizza Delivery

来源：互联网发布：2016淘宝下半年活动编辑：程序博客网时间：2024/05/17 04:11

按照紫书的思路，重点在于可以忽略有哪些具体的位置没有送到，而仅仅关注于已经送到的地方以及下一步需要送到哪些地方，具体实现见如下代码：

#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>#include<cstdio>#include<deque>using namespace std;int T,n;int position[110], value[110];int dp[110][110][2][110];int visit[110][110][2][110];int get(int l,int r,int pos,int amount){if (amount == 0) return 0;if (visit[l][r][pos][amount] == T) return dp[l][r][pos][amount];visit[l][r][pos][amount] = T;int& ans = dp[l][r][pos][amount];ans = 0;for (int i = 0; i < l; i++) ans = max(ans, get(i, r, 0, amount - 1) - amount*abs(position[i] -(pos ? position[r] : position[l])) + value[i]);for (int i = n - 1; i > r; i--) ans = max(ans, get(l, i, 1, amount - 1) - amount*abs(position[i] -(pos ? position[r] : position[l])) + value[i]);return ans;}int main(){cin >> T;memset(visit,0,sizeof(visit));while (T--){cin >> n;for (int i = 0; i < n; i++) cin >> position[i];for (int i = 0; i < n; i++) cin >> value[i];int ans = 0;for (int k = 1; k <= n; k++){for (int i = 0; i < n; i++){ans = max(ans, get(i, i, 0, k - 1) - k*abs(position[i]) + value[i]);}}cout << ans << endl;}}

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