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0MS 代码 两个剪枝#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
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