POJ2718 Smallest Difference 【贪心+枚举】

Smallest Difference

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

1
0 1 2 4 6 7

Sample Output

28

Source

Rocky Mountain 2005

/*
** Problem: POJ2718
** Status: Accepted
** Running Time: 16ms
** Author: Changmu
**
** 题意：给出最多10个单位数字，将这些数字分成两部分，求它们的
** 绝对值的最小值。
** 题解：开始总想着用搜索来枚举，，，后来才知道可以用贪心。具体的思路
** 是这样：如果N为2且有一个数为0时特判，如果N为偶数，那么找出最接近的
** 两个非零数，大的放到box1中，小的放在box2中，剩下的将最小的组合放到
** box1中，最大的组合放在box2，求出差值并更新ans；如果N为奇数，那么在
** 这N个数中，选择最小的一个非零数放到box1头边，然后剩下的数按照N为偶数的情况处理。
*/

#include <stdio.h>
#include <string.h>
#include <algorithm>

#define maxn 12
#define inf 0x3f3f3f3f

int box[maxn], box1[maxn], box2[maxn], vis[maxn], num, num1, num2;
int minSub, ans, sign; // sign 用来标记奇偶

void Select(int k) {
    for(int i = 0; k < num1 && i < num; ++i)
        if(!vis[i]) {
            vis[i] = 1;
            box1[k++] = box[i];
        }
    for(int i = num - 1; sign < num2 && i >= 0; --i)
        if(!vis[i]) {
            box2[sign++] = box[i];
        }
    int a = 0, b = 0;
    for(int i = 0; i < num1; ++i)
        a = a * 10 + box1[i];
    for(int i = 0; i < num2; ++i)
        b = b * 10 + box2[i];
    ans = std::min(ans, a - b);
}

int main() {
    int T, i;
    char ch;
    scanf("%d", &T);
    while(T--) {
        num = 0;
        while(scanf("%d%c", &box[num++], &ch)) {
            vis[num-1] = 0;
            if(ch == '\n') break;
        }

        std::sort(box, box + num);
        num2 = num >> 1;
        num1 = num - num2;
        ans = inf;

        if(num == 2 && box[0] == 0) { // 特判
            printf("%d\n", box[1]);
            continue;
        }

        minSub = 10;
        sign = 0;

        if(num & 1) {
            if(box[0]) {
                vis[0] = 1;
                box1[0] = box[0];
            } else {
                vis[1] = 1;
                box1[0] = box[1];
            }
            Select(1);
        } else {
            i = 1;
            if(!box[0]) i = 2;
            for( ; i < num; ++i)
                minSub = std::min(minSub, box[i] - box[i-1]);
            for(i = 1; i < num; ++i) {
                if(box[i-1] && box[i] - box[i-1] == minSub) {
                    vis[i] = 1; vis[i-1] = 1;
                    box1[0] = box[i];
                    box2[0] = box[i-1];
                    sign = 1;
                    Select(1);
                    memset(vis, 0, sizeof(vis));
                }
            }
        }

        printf("%d\n", ans);
    }
    return 0;
}