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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
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