NBUT1477：AmeriDarts

http://ac.nbutoj.com/Problem/view.xhtml?id=1477

问题描述

AmeriDarts is a dangerous but full of glamour. Mr.cai wants to have a try.
The rule is simple:throw some boomerangs in a line from left to right as quickly as you can. You are asked to throw twice, first time throw A red boomerangs, second time throw B blue boomerangs. Your score is to calculate the shortest length of blue and red boomerang. This rule seems strange? so am I.

输入

First there is two integer A(no more than 1000) and B(no more than 1000), then followed two lines consist of those A and B numbers.Note that these two array both in ascending order.

输出

For each testcase, output the shortest length between red and blue boomerang in a single line.

样例输入

4 4
1 2 3 4
5 6 7 8
5 5
5 12 14 15 16
1 2 7 8 9

样例输出

12

 

题意：找出两个数组中数字差距最小的距离

思路：由于数组是递增的，所以只要记录b数组比较到了那个数即可，无需回溯

 

#include <stdio.h>
#include <algorithm>
using namespace std;

int jian(int a,int b)
{
    int k = a-b;
    if(k<0)
    return -k;
    return k;
}

int main()
{
    int n,m,i,j,l,t;
    int a[1005],b[1005],c[1005];
    while(~scanf("%d%d",&n,&m))
    {
        l = 0;
        for(i = 1;i<=n;i++)
        scanf("%d",&a[i]);
        for(i = 1;i<=m;i++)
        scanf("%d",&b[i]);
        b[0] = 0;
        b[m+1] = 0;
        t = 1;
        for(i = 1;i<=n;i++)
        {
            for(j = t;j<=m;j++)
            {
                if(jian(a[i],b[j])<jian(a[i],b[j-1]) && jian(a[i],b[j])<jian(a[i],b[j+1]))//找到差距最小的地方
                {
                    t = j;//记录b数组所到的位置，放置回溯，节约时间
                    c[l++] = jian(a[i],b[j]);
                }
            }
        }
        sort(c,c+l);
        printf("%d\n",c[0]);
    }

    return 0;
}