P3052 [USACO12MAR]摩天大楼里的奶牛 [状压DP]

传送门

用f[i]记录状态为i的最小值

g[i] 记录状态为i 的剩下电梯的空间的最大值

f[i|1<<j]=f[i](a[j]<=g[i]),g[i|1<<j]=g[i]-a[j]

f[i|1<<j]=f[i]+1(a[j]>g[i]),g[i|1<<j]=w-a[j]

#include<bits/stdc++.h>
#define N 19
using namespace std;
int f[1<<N],g[1<<N];
int n,a[N],w;
int main(){
	scanf("%d%d",&n,&w); memset(f,127,sizeof(f));
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		f[1<<(i-1)] = 1; g[1<<(i-1)] = w-a[i];
	} 
	for(int i=0;i<=(1<<n)-1;i++) for(int j=1;j<=n;j++){
		if((1<<(j-1)) & i) continue;
		if(a[j]<=g[i]){
			f[i|(1<<(j-1))] = min(f[i|(1<<(j-1))],f[i]);
			g[i|(1<<(j-1))] = max(g[i|(1<<(j-1))],g[i]-a[j]);
		} else{
			f[i|(1<<(j-1))] = min(f[i|(1<<(j-1))],f[i]+1);
			g[i|(1<<(j-1))] = max(g[i|(1<<(j-1))],w-a[j]);
		}
	} printf("%d",f[(1<<n)-1]); return 0;
}

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转载自blog.csdn.net/sslz_fsy/article/details/85131744