洛谷 P1541 乌龟棋
题目传送门
解题思路:
一开始脑抽,以为是一道背包,结果写完后发现并不是一道背包.....
f[i][j][x][y]表示1,2,3,4种牌各用了i,j,x,y张可以获得的最大值.
AC代码:
1 #include<iostream> 2 #include<cstdio> 3 #define f(i,j,x,y) f[i][j][x][y] 4 5 using namespace std; 6 7 int n,m,a[351],sum[5],f[41][41][41][41]; 8 9 int main() { 10 scanf("%d%d",&n,&m); 11 for(int i = 1;i <= n; i++) 12 scanf("%d",&a[i]); 13 f(0,0,0,0) = a[1]; 14 for(int i = 1;i <= m; i++) { 15 int u; 16 scanf("%d",&u); 17 sum[u]++; 18 } 19 for(int i = 0;i <= sum[1]; i++) 20 for(int j = 0;j <= sum[2]; j++) 21 for(int x = 0;x <= sum[3]; x++) 22 for(int y = 0;y <= sum[4]; y++) { 23 int now = 1 * i + 2 * j + 3 * x + 4 * y + 1; 24 if(i != 0) f(i,j,x,y) = max(f(i,j,x,y),f(i-1,j,x,y) + a[now]); 25 if(j != 0) f(i,j,x,y) = max(f(i,j,x,y),f(i,j-1,x,y) + a[now]); 26 if(x != 0) f(i,j,x,y) = max(f(i,j,x,y),f(i,j,x-1,y) + a[now]); 27 if(y != 0) f(i,j,x,y) = max(f(i,j,x,y),f(i,j,x,y-1) + a[now]); 28 } 29 printf("%d",f(sum[1],sum[2],sum[3],sum[4])); 30 return 0; 31 }
//NOIP提高2010 T2
