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Re:0MS 代码 两个剪枝In Reply To:0MS 代码 两个剪枝 Posted by:laoyang103 at 2011-08-16 16:55:54 > #include <stdio.h> > #include <stdlib.h> > int data[65] = {0}; > bool findout = false; > int cmp(const void *pa,const void *pb) > { > return *(int *)pb - *(int *)pa; > } > bool dfs(bool flag[],int depth,int n,int sum,int sticks) > { > if(sticks == depth) > { > printf("%d\n",n); > findout = true; > return true; > } > int i,j,k = 0; > int mem = 0; > for(i = 0;i<sticks;i++) > { > if(!flag[i]) > { > if(data[i]+sum <= n && mem != data[i]) > { > mem = data[i]; > flag[i] = true; > if(dfs(flag,depth+1,n,(sum+data[i])%n,sticks)) > return true; > flag[i] = false; > if(data[i] == n-sum || 0 == sum) > return false; > } > } > } > return false; > } > /* > 4 > 4 3 2 1 > 9 > 5 2 1 5 2 1 5 2 1 > 9 > 15 11 8 8 8 4 3 2 1 > 12 > 1 1 2 2 2 3 3 3 3 3 3 4 > 27 > 15 3 2 4 11 1 8 8 8 15 3 2 4 11 1 8 8 8 15 3 2 4 11 1 8 8 8 > 9 > 15 4 3 1 2 8 11 8 8 > */ > int main() > { > int i,j,k; > int n = 0; > while(scanf("%d",&n) && n) > { > bool flag[65] = {0}; > int sum = 0; > for(i = 0;i<n;i++) > { > scanf("%d",&data[i]); > sum+=data[i]; > } > qsort(data,n,4,cmp); > for(i = data[0];;i++) > { > if(0 == sum%i) > { > dfs(flag,0,i,0,n); > if(findout) > { > findout = false; > break; > } > } > } > } > return 0; > } > > 以下剪枝的前提条件为搜索失败: > 1.如果当前搜索的长度和当前要组合成的木棒尚缺的长度相等,也就是说当前搜索的长度+当前累计的长度 = 要搜索的长度,那么剪枝。原因为这个长度不放这里,那他也不能放别处。 > 2.如果当前木棒累计长度为0,那么剪枝,原因为子问题无解,整个问题也不会有解 看上去大部分内容是参考上一位贴代码的同学的成果 Followed by: Post your reply here: |
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