洛谷 P2979 [USACO10JAN]奶酪塔Cheese Towers
洛谷 P2979 [USACO10JAN]奶酪塔Cheese Towers
分两种状态 一种最上面有大奶酪
一种没有
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <iostream> 8 #include <iomanip> 9 #define For(i,j,k) for(int i=j;i<=k;i++) 10 #define Dow(i,j,k) for(int i=j;i>=k;i--) 11 #define LL long long 12 using namespace std ; 13 inline int read() { 14 int x = 0 , f = 1 ; 15 char ch = getchar() ; 16 while(ch<'0'||ch>'9') { if(ch=='-') f = -1 ; ch = getchar() ; } 17 while(ch>='0'&&ch<='9') { x = x * 10+ch-48 ; ch = getchar() ; } 18 return x * f ; 19 } 20 21 const int N = 1011, inf = 1e9; 22 int n,H,K,ans,tot; 23 int f[N][2]; 24 struct node{ 25 int h,v; 26 }a[111]; 27 inline bool cmp_1(node a, node b) { return a.h > b.h; } 28 inline bool cmp_2(node a, node b) { return a.h < b.h; } 29 30 int main() { 31 n=read(); H=read(); K=read(); 32 For(i, 1, n) a[i].v=read(), a[i].h = read(); 33 For(i, 1, H) f[i][1]=-1; 34 sort(a+1, a+n+1, cmp_1); 35 for(; a[tot+1].h >= K; tot++) 36 sort(a+1, a+tot+1, cmp_2); 37 38 For(i, 1, n) 39 For(j, a[i].h, H) { 40 f[j][0] = max(f[j][0], f[j-a[i].h][0]+a[i].v); 41 if(f[j-a[i].h/5*4][1]!=-1) f[j][1] = max(f[j][1], f[j-a[i].h/5*4][1]+a[i].v); 42 if(a[i].h>=K) f[j][1] = max(f[j][1], f[(j-a[i].h)/5*4][0]+a[i].v ); 43 } 44 printf("%d\n",max(f[H][0], f[H][1])); 45 return 0; 46 }
