标签：print names needed ber rate follow 描述 single 状压
给出n个物品，体积为w[i]，现把其分成若干组，要求每组总体积<=W，问最小分组。(n<=18)
输入格式：
Line 1: N and W separated by a space.
输出格式：
Line 1: A single integer, R, indicating the minimum number of elevator rides needed.
one of the R trips down the elevator. Each line starts with an integer giving the number of cows in the set, followed by the indices of the individual cows in the set.
见代码：
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 const int inf=0x3f3f3f3f; 6 int n,v,w[18],end; 7 int dp[20][1<<20]; 8 int main() { 9 /* 10 此处状态为前i个电梯，（bit)j的最小重量 11 dp[i][j]=>dp[i][j|(1<<k)]||dp[i+1][j|(1<<k)] ,!(k&(1<<k)) 12 */ 13 scanf("%d%d",&n,&v),end=1<<n; 14 for(int i=0;i<n;i++) scanf("%d",&w[i]); 15 for(int i=0;i<n;i++) for(int j=0;j<end;j++) dp[i][j]=inf; 16 for(int i=0;i<n;i++) dp[1][1<<i]=w[i]; 17 for(int i=0;i<=n;i++) for(int j=0;j<end;j++) if(dp[i][j]!=inf)//存在 18 for(int k=0;k<n;k++) if(!(j&(1<<k)))//不存在 19 if(dp[i][j]+w[k]<=v) dp[i][j|(1<<k)]=min(dp[i][j|(1<<k)],dp[i][j]+w[k]); 20 else dp[i+1][j|(1<<k)]=min(dp[i][j|(1<<k)],w[k]); 21 for(int i=0;i<=n;i++) if(dp[i][end-1]!=inf) { 22 printf("%d\n",i); 23 break; 24 } 25 return 0; 26 }
P3052 [USACO12MAR]摩天大楼里的奶牛Cows in a Skyscraper
标签：print names needed ber rate follow 描述 single 状压
原文地址：http://www.cnblogs.com/InfoEoR/p/7881504.html
