poj 2718 Smallest Difference 【dfs(全排列变形题)】

Smallest Difference

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6389 Accepted: 1736

Description

Given a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing them in some order. The remaining digits can be written down in some order to form a second integer. Unless the resulting integer is 0, the integer may not start with the digit 0. 

For example, if you are given the digits 0, 1, 2, 4, 6 and 7, you can write the pair of integers 10 and 2467. Of course, there are many ways to form such pairs of integers: 210 and 764, 204 and 176, etc. The absolute value of the difference between the integers in the last pair is 28, and it turns out that no other pair formed by the rules above can achieve a smaller difference.

Input

The first line of input contains the number of cases to follow. For each case, there is one line of input containing at least two but no more than 10 decimal digits. (The decimal digits are 0, 1, ..., 9.) No digit appears more than once in one line of the input. The digits will appear in increasing order, separated by exactly one blank space.

Output

For each test case, write on a single line the smallest absolute difference of two integers that can be written from the given digits as described by the rules above.

Sample Input

10 1 2 4 6 7

Sample Output

28

Source

Rocky Mountain 2005

思路：

         通过全排列就能将两个数的所有可能都排出来，然后取前一半数，来组成一个数，后一半数都保存到另一个数，然后做差就行了！不管是奇数个数还是偶数个数，都取n/2个数就行！

          但是要特殊处理，如果是输入两个数（如：0 1），就会出错，所以要特殊处理，而如果输入10 个数，那么会造成超时（看讨论区说超时，但是试了试没超时，所以不用特判），所以将两个边界特殊考虑一下就OK了！

代码：

#include <stdio.h>#include <string.h>#include <algorithm>#define INF 0x3f3f3f3fusing namespace std;char a[25];int b[15];int used[15];int perm[15];int c[15];int num;int minn;int u,v;int vis[15];int perm2[15];void permutation(int pos,int n){if(pos==n){u=0,v=0;for(int i=0;i<n/2;i++){u=u*10+perm[i];}for(int i=n/2;i<n;i++){v=v*10+perm[i];}int dif=u-v;if(dif<0)dif=-dif;minn=minn>dif?dif:minn;return;}for(int i=0;i<num;i++){if(used[i])continue;if((pos==0||pos==num/2)&&b[i]==0)//pos<num-1&&(可以加上这一句，能够将0，1这个例子包含进去)continue;perm[pos]=b[i];used[i]=1;permutation(pos+1,n);used[i]=0;}}int main(){//freopen("shuju.txt","r",stdin);int T;scanf("%d",&T);getchar();//这个不能写到下面否则会出错！ while(T--){gets(a);int len=strlen(a);int j=0;for(int i=0;i<len;i++){if(a[i]>='0'&&a[i]<='9'){b[j++]=a[i]-'0';//从小到大 }}num=j;if(num==2)//注意两个数字的时候，直接相减就行了！（如果不加这一句输入0，1就会出错！） {int s=b[0]-b[1];if(s<0)s=-s;printf("%d\n",s);continue;}if(num==10){printf("247\n");continue;}memset(used,0,sizeof(used));minn=INF;permutation(0,num);        printf("%d\n",minn);}return 0;}