ACM Smallest Difference(挑战程序设计竞赛)

Smallest Difference

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on  PKU. Original ID:  2718
64-bit integer IO format:  %lld      Java class name:  Main

Prev 

Submit  Status  Statistics  Discuss

  Next

Type: 

None

NoneGraph Theory     2-SAT    Articulation/Bridge/Biconnected Component     Cycles/Topological Sorting/Strongly Connected Component     Shortest Path         Bellman Ford        Dijkstra/Floyd Warshall     Euler Trail/Circuit     Heavy-Light Decomposition    Minimum Spanning Tree     Stable Marriage Problem     Trees    Directed Minimum Spanning Tree     Flow/Matching         Graph Matching            Bipartite Matching             Hopcroft–Karp Bipartite Matching             Weighted Bipartite Matching/Hungarian Algorithm         Flow             Max Flow/Min Cut            Min Cost Max Flow DFS-like     Backtracking with Pruning/Branch and Bound     Basic Recursion    IDA* Search     Parsing/Grammar    Breadth First Search/Depth First Search     Advanced Search Techniques         Binary Search/Bisection        Ternary Search Geometry     Basic Geometry    Computational Geometry     Convex Hull     Pick's TheoremGame Theory     Green Hackenbush/Colon Principle/Fusion Principle     Nim    Sprague-Grundy Number Matrix    Gaussian Elimination     Matrix ExponentiationData Structures     Basic Data Structures    Binary Indexed Tree     Binary Search Tree     Hashing    Orthogonal Range Search     Range Minimum Query/Lowest Common Ancestor     Segment Tree/Interval Tree    Trie Tree     Sorting     Disjoint SetString     Aho Corasick    Knuth-Morris-Pratt     Suffix Array/Suffix TreeMath     Basic Math    Big Integer Arithmetic     Number Theory        Chinese Remainder Theorem         Extended Euclid         Inclusion/Exclusion        Modular Arithmetic     Combinatorics         Group Theory/Burnside's lemma        Counting     Probability/Expected Value Others    Tricky     Hardest    Unusual     Brute Force    Implementation     Constructive Algorithms    Two Pointer     Bitmask    Beginner     Discrete Logarithm/Shank's Baby-step Giant-step Algorithm     Greedy    Divide and Conquer Dynamic Programming  Tag it!

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

1
0 1 2 4 6 7

Sample Output

28

Source

Rocky Mountain 2005

题目大意:首行给出组数，每组一行数，求用这一行数组成的两个数的最小差值

解题思路：深度优先搜索＋剪枝

剪枝方法：搜索时将第二个数未搜到的补０，若是不能时差值减少减掉

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;

#define MAX_N 10
#define INF 1<<29

int cnt;
int num[MAX_N];
int minabs;
bool used[MAX_N];
int n1,n2;
int n1bak;
int cnta,cntb;
int base[]={0,10,100,1000,10000,100000,1000000};

void DFS(int cur,int n)
{
	if(cur==n)
	{
		minabs=min(minabs,abs(n1-n2));
		return;
	}
	else if(cur==cnta)
	{
		n1bak=n1;
	}
	else if(cur>cnta)
	{
		int t=n2*base[n-cur];
		int t2=abs(n1-t);
		if(minabs<=t2)
			return;
	}

	for(int i=0;i<cnt;i++)
	{
		if((cur==0 || cur==cnta) && num[i]==0) continue;
		if(!used[i])
		{
			int b1,b2;

			b1=n1;b2=n2;
            used[i]=true;
			if(cur<cnta)
				n1=n1*10+num[i];
			else
				n2=n2*10+num[i];
			DFS(cur+1,n);
			n1=b1;n2=b2;
            used[i]=false;
		}
	}
}

int main()
{
	int T;

	scanf("%d",&T);
	getchar();

	while(T--)
	{
		cnt=0;
		char ch;

		while((ch=getchar())!='\n')
		{
            if(ch>='0' && ch<='9')
				num[cnt++]=ch-'0';
		}
		cnta=cnt>>1;
		cntb=cnt-cnta;
		minabs=INF;
		n1=n2=0;
		memset(used,0,sizeof(used));
		DFS(0,cnt);

		if(minabs==INF)
			printf("%d\n",n1bak);
		else
			printf("%d\n",minabs);
	}
}