2015 多校联赛 ——HDU5353（构造）

最新推荐文章于 2019-08-10 14:18:00 发布

原创最新推荐文章于 2019-08-10 14:18:00 发布 · 482 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#hdu #2015联赛 #构造

ACM/ICPC 专栏收录该内容

67 篇文章

订阅专栏

本文讨论了如何通过交换或转移糖果使相邻两个数的糖果数量相等的问题，并提供了算法实现。通过枚举和操作数组元素，最终确定是否能够使所有数的糖果数量一致。若能，则提供解决方案；否则输出失败。

Each soda has some candies in their hand. And they want to make the number of candies the same by doing some taking and giving operations. More specifically, every two adjacent soda x and y can do one of the following operations only once:
1. x-th soda gives y-th soda a candy if he has one;
2. y-th soda gives x-th soda a candy if he has one;
3. they just do nothing.

Sample Input

3
6
1 0 1 0 0 0
5
1 1 1 1 1
3
1 2 3

Sample Output

NO
YES
0
YES
2
2 1
3 2

对相邻两个数之间进行以下三种操作的一种,最后使他们相等

①a++ b-- ②a-- b++ ③nothing

如果(sum%n != 0)，直接失败。用一个数组来记录数与平均数之间的差值。

先枚举第一位数的三种情况，

当C[i] == 1时，从C[i+1]取一；

当C[i] == -1时，给C[i+1]一个；

当C[i] == 0时，nothing；

else：false。

ps：完全没想到要对第一位进行枚举 OoO，一直wa

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long ll;
const int mod = 1000000007;
int a[100050];
int aver,flag,n;
int p[100050][2];
int c[100050];
int tot;

bool solve()
{
    for(int i = 3; i <= n; i++)
        c[i] = a[i];
    for(int i = 2; i <= n; i++)
    {
        if(c[i] == 0)
            continue;
        else if(c[i] == 1 && i != n)
        {
            c[i+1]++;
            c[i]--;
            p[tot][0] = i;
            p[tot++][1] = i+1;
        }
        else if(c[i] == 1 && i == n)
        {
            c[1]++;
            c[i]--;
            p[tot][0] = i;
            p[tot++][1] = 1;
        }
        else if(c[i] == -1 && i!= n)
        {
            c[i]++;
            c[i+1]--;
            p[tot][0] = i+1;
            p[tot++][1] = i;
        }
        else if(c[i] == -1 && i== n)
        {
            c[i]++;
            c[1]--;
            p[tot][0] = 1;
            p[tot++][1] = i;
        }
        else if(c[i] >1 || c[i] < -1)
            return false;
    }
    for(int i = 1; i <= n; i++)
        if(c[i]!=0)
            return false;
    return true;
}

void prin()
{
    printf("YES\n");
    printf("%d\n",tot);
    for(int i = 0; i < tot; i++)
        printf("%d %d\n",p[i][0],p[i][1]);
}

int main()
{
    int T;
    //freopen("01.txt","r",stdin);
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        ll sum = 0;
        memset(p,0,sizeof(p));
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&a[i]);
            sum += a[i];
        }

        if(sum % n)
        {
            printf("NO\n");
            continue;
        }
        aver = sum / n;
        flag = 0;
        tot = 0;
        for(int j = 1; j <= n; j++)
            a[j] =a[j] - aver;
        c[1] = a[1];
        c[2] = a[2];
        if(solve())
        {
            prin();
        }
        else
        {
            tot = 0;
            c[1] = a[1] - 1;
            c[2] = a[2] + 1;
            p[tot][0] = 1;
            p[tot++][1] = 2;
            if(solve())
                prin();
            else
            {
                tot = 0;
                c[1] = a[1] + 1;
                c[2] = a[2] - 1;
                p[tot][0] = 2;
                p[tot++][1] = 1;
                if(solve())
                    prin();
                else
                    printf("NO\n");
            }
        }

    }
    return 0;
}