参照题解
题目本质：最优决策一定只有两种：前X大的A值、前X-1大的A值加上一个A+2*S最大的。
解决方法：
按照A的从大到小排序。
维护：1.A的前缀和；2.前i个里最大的S；3.从i往后最大的A+2*S.
然后O（n）max一遍即可。
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <cmath> 6 #include <ctime> 7 #include <cctype> 8 #include <climits> 9 #include <iostream> 10 #include <iomanip> 11 #include <algorithm> 12 #include <string> 13 #include <sstream> 14 #include <stack> 15 #include <queue> 16 #include <set> 17 #include <map> 18 #include <vector> 19 #include <list> 20 #include <fstream> 21 #define ri readint() 22 #define gc getchar() 23 #define R(x) scanf("%d", &x) 24 #define W(x) printf("%d ", x) 25 #define init(a, b) memset(a, b, sizeof(a)) 26 #define rep(i, a, b) for (int i = a; i <= b; i++) 27 #define irep(i, a, b) for (int i = a; i >= b; i--) 28 #define ls p << 1 29 #define rs p << 1 | 1 30 using namespace std; 31 32 typedef double db; 33 typedef long long ll; 34 typedef unsigned long long ull; 35 typedef pair<int, int> P; 36 const int inf = 0x3f3f3f3f; 37 const ll INF = 1e18; 38 39 inline int readint() { 40 int x = 0, s = 1, c = gc; 41 while (c <= 32) c = gc; 42 if (c == '-') s = -1, c = gc; 43 for (; isdigit(c); c = gc) 44 x = x * 10 + c - 48; 45 return x * s; 46 } 47 48 const int maxn = 1e5 + 5; 49 struct family { 50 int s, a; 51 52 family() { 53 s = a = inf; 54 } 55 56 bool operator < (const family b) const { 57 return a > b.a; 58 } 59 }; 60 61 int main() { 62 int n = ri; 63 vector<family> v(n + 1); 64 rep(i, 1, n) v[i].s = ri; 65 rep(i, 1, n) v[i].a = ri; 66 sort(v.begin(), v.end()); 67 68 int maxpres[n+1], suma[n+1], maxsufall[n+2]; 69 init(maxpres, 0), init(suma, 0), init(maxsufall, 0); 70 rep(i, 1, n) maxpres[i] = max(maxpres[i - 1], v[i].s); 71 rep(i, 1, n) suma[i] = suma[i - 1] + v[i].a; 72 irep(i, n, 1) maxsufall[i] = max(maxsufall[i + 1], v[i].a + v[i].s * 2); 73 74 vector<int> ans; 75 rep(i, 1, n) ans.push_back(max(suma[i] + 2 * maxpres[i], suma[i - 1] + maxsufall[i])); 76 for (int i : ans) W(i); 77 return 0; 78 }