思路:dp[i][j][k],第一维表示用钱买,第二维表示用积分换,第三维表示免费的。对于每个商品而言,要么用钱,要么用积分,要么用免费的次数。
所以呢,dp[i][j][k] = max(dp[i - a[s]][j][k],dp[i][j - b[s]][k],dp[i][j][k - 1]) + v[s];这个是对于地s个物品而言的。
/***************************************** Author :Crazy_AC(JamesQi) Time :2015 File Name : *****************************************/ // #pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <algorithm> #include <iomanip> #include <sstream> #include <string> #include <stack> #include <queue> #include <deque> #include <vector> #include <map> #include <set> #include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> #include <limits.h> using namespace std; #define MEM(a,b) memset(a,b,sizeof a) typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> ii; const int inf = 1 << 30; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; inline int Readint(){ char c = getchar(); while(!isdigit(c)) c = getchar(); int x = 0; while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); } return x; } const int maxn = 110; int dp[110][110][6];//qian,jifen,mianfei int a[maxn],b[maxn],v[maxn]; int n,v1,v2,k; int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); while(~scanf("%d%d%d%d",&n,&v1,&v2,&k)){ for (int i = 0;i < n;i++) scanf("%d%d%d",&a[i],&b[i],&v[i]); memset(dp, 0,sizeof dp); int ans = 0; for (int i = 0;i < n;i++){ for (int x1 = v1;x1 >= 0;x1--){ for (int x2 = v2;x2 >= 0;x2--){ for (int x3 = k;x3 >= 0;x3--){ int tmp = 0; if (x1 >= a[i]) tmp = max(tmp,dp[x1 - a[i]][x2][x3] + v[i]); if (x2 >= b[i]) tmp = max(tmp,dp[x1][x2 - b[i]][x3] + v[i]); if (x3 > 0) tmp = max(tmp,dp[x1][x2][x3 - 1] + v[i]); // ans = max(ans,dp[x1][x2][x3]); dp[x1][x2][x3] = max(dp[x1][x2][x3],tmp); } } } } // cout << "ans = " << ans << '\n'; printf("%d\n",dp[v1][v2][k]); } return 0; }