HDU 6188 Duizi and Shunzi（贪心）

青旅半醒 2022-06-09 11:56 87阅读 0赞

Nike likes playing cards and makes a problem of it.  
  
Now give you n integers, ai(1≤i≤n) ai(1≤i≤n)  
  
We define two identical numbers (eg: 2,2 2,2) a Duizi,  
and three consecutive positive integers (eg: 2,3,4 2,3,4) a Shunzi.  
  
Now you want to use these integers to form Shunzi and Duizi as many as possible.  
  
Let s be the total number of the Shunzi and the Duizi you formed.  
  
Try to calculate max(s) max(s).  
  
Each number can be used only once.

Input

The input contains several test cases.  
  
For each test case, the first line contains one integer n( 1≤n≤106 1≤n≤106).  
Then the next line contains n space-separated integers ai ai ( 1≤ai≤n 1≤ai≤n)

Output

For each test case, output the answer in a line.

Sample Input

7
    1 2 3 4 5 6 7
    9
    1 1 1 2 2 2 3 3 3
    6
    2 2 3 3 3 3 
    6
    1 2 3 3 4 5

Sample Output

2
    4
    3
    2

Hint

Case 1（1，2，3）（4，5，6）
    
    Case 2（1，2，3）（1，1）（2，2）（3，3）
    
    Case 3（2，2）（3，3）（3，3）
    
    Case 4（1，2，3）（3，4，5）

题解：

题意：

给你一堆牌，让你求出最大的对子数和顺子数（3个连续的牌）的和

思路：

从右到左扫一遍牌，先把对子打掉，如果还剩下牌，就看下一个数字的牌数是否为奇数，如果是并且下下张牌是有牌的，就连成顺子打出去，这样贪心思路其实是用“废弃"的当前和废弃的下一个奇数牌，使得消耗一个有用的下下张牌就能换成一对，而如果是只打下下张牌的对子那么就是消耗了2张有用的牌（暂时判定）来换来一个对子

代码：

#include<iostream>
    #include<cstring>
    #include<stdio.h>
    #include<math.h>
    #include<string>
    #include<stdio.h>
    #include<queue>
    #include<stack>
    #include<map>
    #include<vector>
    #include<deque>
    #include<algorithm>
    using namespace std;
    #define INF 100861111
    #define ll long long
    #define eps 1e-7
    #define maxn 1e5+5
    #define lson k*2
    #define rson k*2+1
    #define M (t[k].l+t[k].r)/2
    int a[1000005];
    int main()
    {
        int i,j,n,ans,x;
        while(scanf("%d",&n)!=EOF)
        {
            ans=0;
            memset(a,0,sizeof(a));
            for(i=1;i<=n;i++)
            {
                scanf("%d",&x);
                a[x]++;
            }
            for(i=1;i<=n;i++)
            {
                ans+=(a[i]/2);
                a[i]%=2;
                if(a[i]&&a[i+1]%2&&a[i+2])
                {
                    ans++;
                    a[i]--;
                    a[i+1]--;
                    a[i+2]--;
                }
            }
            printf("%d\n",ans);
        }
        return 0;
    }