感觉是很巧妙的dp啊, 想到了分成两种信用卡, 并且有一种需要按b排序, 但是没想出怎么dp。。
#include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 500 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;} int n; LL dp[N][N]; struct Node { int a, b, k; void read() { scanf("%d%d%d", &a, &b, &k); } bool operator < (const Node &rhs) const { return b > rhs.b; } } offers[N]; int main() { scanf("%d", &n); for(int i = 1; i <= n; i++) offers[i].read(); sort(offers + 1, offers + 1 + n); LL ans = 0; memset(dp, 0xc0, sizeof(dp)); dp[0][0] = 0; for(int i = 1; i <= n; i++) { for(int j = 0; j <= i; j++) { chkmax(dp[i][j], dp[i - 1][j]); chkmax(dp[i][j], dp[i - 1][j] + offers[i].a - 1LL * offers[i].k * offers[i].b); if(j) chkmax(dp[i][j], dp[i - 1][j - 1] + offers[i].a - 1LL * (j - 1) * offers[i].b); chkmax(ans, dp[i][j]); } } printf("%lld ", ans); return 0; } /* */