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40行状态压缩DFS水过... 1秒..状态压缩记录哪些棍子用过了 一共4层每层从1<<start搜到(1<<n)-1.. 符合要求就记录下哪些用过了然后进入下一层.. #include <stdio.h> bool dfs(int a[],int u,int n,int __s,int start) { if (start==n) return true; for(int i=(1<<start);i<(1<<n);i+=(2<<start)) { if(i&u) continue; int count=0; for(int j=0;j<n;j++) count+=a[j]*((i>>j)&1); if(count!=__s) continue; int __u=u|i; int j; for(j=0;j<n && ((__u>>j)&1);j++); if(dfs(a,__u,n,__s,j)) { return true; } } return false; } int main() { int N; scanf("%d",&N); while (N--) { int n,a[21],count=0; scanf("%d",&n); for (int i=0;i<n;i++) { scanf("%d",a+i); count+=a[i]; } printf("%s\n",count%4?"no":(dfs(a,0,n,count/4,0)?"yes":"no")); } return 0; } Followed by: Post your reply here: |
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