浅谈枚举

枚举谁都会，但是同样都是打枚举，得分就可能不一样，就比如砝码称重那道题，神仙涛枚举100分，我枚举30分,这就是枚举技巧，所以对于我这种算法基本靠枚举的蒟蒻来说学好枚举很重要了...咕咕咕

定义

枚举就是需要遍历每个解来寻找最优解/计数的问题 ，复杂度会出现指数级，此时数据范围一般较小 关键在于能否找到需要枚举的变量，以及枚举的效率，避免无用的枚举

题目链接：

P1036 选数

思路：

这道题就是个简单的搜索或者说是枚举，主要是想用这道题来浅谈一下枚举的技巧，当然每道题都不一样。

最暴力的枚举：

来自题解

#include<stdio.h>
#include<cmath>
#include<cstdlib>
#include<iostream>
using namespace std;
int a[21];
int s,x,n,k;
bool zs(long long y)//判断是否是质数
{
    if (y==1||!y) return 0;
    for (int i=2;i<=sqrt(y);i++)
     if (!(y%i)) return 0;
    return 1;

}
void sr()
{
    scanf("%d %d",&n,&k);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        x+=a[i];//求下总和
    }
}
void js()
{
    if (k==19||k==n-1) {for (int i=1;i<=n;i++) if (zs(x-a[i])) s++;return;}
    if (k==20||n==k) {if (zs(x)) s++;return;}//两个特判
    if (k==1)  
     {
      for (int i=1;i<=n;i++) 
       if (zs(a[i])) s++;//计算
     }
    if (k==2)  
     {
      for (int i=1;i<=n-1;i++) 
       for (int i1=i+1;i1<=n;i1++) 
        if (zs(a[i]+a[i1])) s++;//计算
        return;
     }
    if (k==3)  
     {
      for (int i=1;i<=n-2;i++) 
       for (int i1=i+1;i1<=n-1;i1++) 
        for (int i2=i1+1;i2<=n;i2++) 
         if(zs(a[i]+a[i1]+a[i2]))s++;//计算
         return;
     }
    if (k==4)  
     {
       for (int i=1;i<=n-3;i++) 
        for (int i1=i+1;i1<=n-2;i1++) 
         for (int i2=i1+1;i2<=n-1;i2++) 
          for (int i3=i2+1;i3<=n;i3++)
          if(zs(a[i]+a[i1]+a[i2]+a[i3]))s++;
          return;
     }
    if (k==5)  
     {
       for (int i=1;i<=n-4;i++) 
        for (int i1=i+1;i1<=n-3;i1++) 
         for (int i2=i1+1;i2<=n-2;i2++) 
          for (int i3=i2+1;i3<=n-1;i3++)
           for (int i4=i3+1;i4<=n;i4++)
          if(zs(a[i]+a[i1]+a[i2]+a[i3]+a[i4]))s++;
          return;
     }
    if (k==6)
     {
       for (int i=1;i<=n-5;i++)
        for (int i1=i+1;i1<=n-4;i1++) 
         for (int i2=i1+1;i2<=n-3;i2++) 
          for (int i3=i2+1;i3<=n-2;i3++)
           for (int i4=i3+1;i4<=n-1;i4++)
            for (int i5=i4+1;i5<=n;i5++)
          if(zs(a[i]+a[i1]+a[i2]+a[i3]+a[i4]+a[i5]))s++;
          return;
     }
    if (k==7)
     {
       for (int i=1;i<=n-6;i++) 
        for (int i1=i+1;i1<=n-5;i1++)
         for (int i2=i1+1;i2<=n-4;i2++) 
          for (int i3=i2+1;i3<=n-3;i3++)
           for (int i4=i3+1;i4<=n-2;i4++)
            for (int i5=i4+1;i5<=n-1;i5++)
             for (int i6=i5+1;i6<=n;i6++)
          if(zs(a[i]+a[i1]+a[i2]+a[i3]+a[i4]+a[i5]+a[i6]))s++;
          return;
     }
    if (k==8)  
     {
       for (int i=1;i<=n-7;i++) 
        for (int i1=i+1;i1<=n-6;i1++) 
         for (int i2=i1+1;i2<=n-5;i2++) 
          for (int i3=i2+1;i3<=n-4;i3++)
           for (int i4=i3+1;i4<=n-3;i4++)
            for (int i5=i4+1;i5<=n-2;i5++)
             for (int i6=i5+1;i6<=n-1;i6++)
              for (int i7=i6+1;i7<=n;i7++)
          if(zs(a[i]+a[i1]+a[i2]+a[i3]+a[i4]+a[i5]+a[i6]+a[i7]))s++;
          return;
     }
    if (k==9)
     {
       for (int i=1;i<=n-8;i++) 
        for (int i1=i+1;i1<=n-7;i1++) 
         for (int i2=i1+1;i2<=n-6;i2++) 
          for (int i3=i2+1;i3<=n-5;i3++)
           for (int i4=i3+1;i4<=n-4;i4++)
            for (int i5=i4+1;i5<=n-3;i5++)
             for (int i6=i5+1;i6<=n-2;i6++)
              for (int i7=i6+1;i7<=n-1;i7++)
               for (int i8=i7+1;i8<=n;i8++)
          if(zs(a[i]+a[i1]+a[i2]+a[i3]+a[i4]+a[i5]+a[i6]+a[i7]+a[i8]))s++;
          return;
     }
    if (k==10)
     {
       for (int i=1;i<=n-9;i++) 
        for (int i1=i+1;i1<=n-8;i1++) 
         for (int i2=i1+1;i2<=n-7;i2++) 
          for (int i3=i2+1;i3<=n-6;i3++)
           for (int i4=i3+1;i4<=n-5;i4++)
            for (int i5=i4+1;i5<=n-4;i5++)
             for (int i6=i5+1;i6<=n-3;i6++)
              for (int i7=i6+1;i7<=n-2;i7++)
               for (int i8=i7+1;i8<=n-1;i8++)
                for (int i9=i8+1;i9<=n;i9++)
          if(zs(a[i]+a[i1]+a[i2]+a[i3]+a[i4]+a[i5]+a[i6]+a[i7]+a[i8]+a[i9]))s++;
          return;
     }
    if (k==11)
     {
       for (int i=1;i<=n-8;i++) 
        for (int i1=i+1;i1<=n-7;i1++) 
         for (int i2=i1+1;i2<=n-6;i2++) 
          for (int i3=i2+1;i3<=n-5;i3++)
           for (int i4=i3+1;i4<=n-4;i4++)
            for (int i5=i4+1;i5<=n-3;i5++)
             for (int i6=i5+1;i6<=n-2;i6++)
              for (int i7=i6+1;i7<=n-1;i7++)
               for (int i8=i7+1;i8<=n;i8++)
          if(zs(x-(a[i]+a[i1]+a[i2]+a[i3]+a[i4]+a[i5]+a[i6]+a[i7]+a[i8])))s++;//注意这里是用x去减
          return;
     }
    if (k==12)
     {
       for (int i=1;i<=n-7;i++) 
        for (int i1=i+1;i1<=n-6;i1++) 
         for (int i2=i1+1;i2<=n-5;i2++) 
          for (int i3=i2+1;i3<=n-4;i3++)
           for (int i4=i3+1;i4<=n-3;i4++)
            for (int i5=i4+1;i5<=n-2;i5++)
             for (int i6=i5+1;i6<=n-1;i6++)
              for (int i7=i6+1;i7<=n;i7++)
          if(zs(x-(a[i]+a[i1]+a[i2]+a[i3]+a[i4]+a[i5]+a[i6]+a[i7])))s++;
          return ;//以下都是用x去减，注意！
     }
    if (k==13)
     {
       for (int i=1;i<=n-6;i++) 
        for (int i1=i+1;i1<=n-5;i1++)
         for (int i2=i1+1;i2<=n-4;i2++) 
          for (int i3=i2+1;i3<=n-3;i3++)
           for (int i4=i3+1;i4<=n-2;i4++)
            for (int i5=i4+1;i5<=n-1;i5++)
             for (int i6=i5+1;i6<=n;i6++)
          if(zs(x-(a[i]+a[i1]+a[i2]+a[i3]+a[i4]+a[i5]+a[i6])))s++;
          return;
     }
    if (k==14)
     {
       for (int i=1;i<=n-5;i++)
        for (int i1=i+1;i1<=n-4;i1++) 
         for (int i2=i1+1;i2<=n-3;i2++) 
          for (int i3=i2+1;i3<=n-2;i3++)
           for (int i4=i3+1;i4<=n-1;i4++)
            for (int i5=i4+1;i5<=n;i5++)
          if(zs(x-(a[i]+a[i1]+a[i2]+a[i3]+a[i4]+a[i5])))s++;
          return;
     }
    if (k==15)
     {
       for (int i=1;i<=n-4;i++) 
        for (int i1=i+1;i1<=n-3;i1++) 
         for (int i2=i1+1;i2<=n-2;i2++) 
          for (int i3=i2+1;i3<=n-1;i3++)
           for (int i4=i3+1;i4<=n;i4++)
          if(zs(x-(a[i]+a[i1]+a[i2]+a[i3]+a[i4])))s++;
          return;
     }
    if (k==16)
     {
       for (int i=1;i<=n-3;i++) 
        for (int i1=i+1;i1<=n-2;i1++) 
         for (int i2=i1+1;i2<=n-1;i2++) 
          for (int i3=i2+1;i3<=n;i3++)
          if(zs(x-(a[i]+a[i1]+a[i2]+a[i3])))s++;
          return;
     }
    if (k==17)
     {
       for (int i=1;i<=n-2;i++) 
        for (int i1=i+1;i1<=n-1;i1++) 
         for (int i2=i1+1;i2<=n;i2++) 
          if(zs(x-(a[i]+a[i1]+a[i2])))s++;
          return;
     }
    if (k==18)
     {
       for (int i=1;i<=n-1;i++) 
        for (int i1=i+1;i1<=n;i1++) 
         if (zs(x-(a[i]+a[i1]))) s++;//之前没有用x去减，给大家造成了一些困扰，在这里说声抱歉
         return;
     }
}
int main()
{
   sr();
   js();
   printf("%d",s);
}

这方法说实话没谁了

dfs枚举法：

依次考虑每个元素是否选取 记录当前状态选取元素个数，以及被选取的元素之和 向下一层搜索时，更新状态，搜索结束时，还原状态

/*
    原始的dfs
*/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<string>
#include<cstring>
#define MAXN 100000
using namespace std;
inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * f;
}
int ans;
int a[MAXN];
int n, k;
int is_prime (int x)
{
	if(x==1) return 0;
	for(int i=2;i*i<=x;++i)
	{
		if(x%i==0) return 0;
	}
	return 1;
}
void dfs(int now, int cnt, int s) {
    if (now == n) {
        if (cnt == k && is_prime(s)) ans++;
    } else {
        dfs(now + 1, cnt, s);
        dfs(now + 1, cnt + 1, s + a[now]);
    }
}

dfs(0, 0, 0);

int main() {
	cin>>n>>k;
	for(int i=0;i<n;++i)
	{
		cin>>a[i];
	}
	dfs(0, 0, 0);
	cout<<ans;
	return 0;
}

观察上面的程序发现还是有问题的
它仍然会搜索一些多余状态

所以我们要剪枝

剪枝版dfs

剪枝如下，很显然是吧

if (cnt > k || cnt + n - now < k) return;

代码就不放了，太长了...只放核心：

void dfs(int now, int cnt, int s) {
    if (now == n) {
        if (cnt == k && is_prime(s)) ans++;
    } else {
    	if (cnt > k || cnt + n - now < k) return;//这里**** 
        dfs(now + 1, cnt, s);
        dfs(now + 1, cnt + 1, s + a[now]);
    }
}

dfs枚举排列

依次考虑每个位置选取的元素 记录当前状态已经被选取的元素 向下一层搜索时更新状态，搜索结束时，还原状态

void dfs(int now) {
    if (now == n) {
        if (test(p)) ans++;
    } else {
        for (int i = 1; i <= n; i++) {
            if (!flag[i]) {
                p[now] = i;
                flag[i] = true;
                dfs(now + 1);
                flag[i] = false;
            }
        }
    }
}
dfs(0);

高效枚举集合

这个方法就比较牛逼了

一个集合的子集个数是(2^n)个，而且，我们可以建立一个从(0)到(2^n-1)的整数到集合子集之间的一一映射 具体来说，将最低位记作第(0)位，二进制表示第i位为(1)，则表示取集合中的第(i)个元素 因此，只需枚举(0)到(2^n-1)之间的整数即可 而判断某个元素有没有被选中，就是判断第(i)位是否位(a)

for (int j = 0; j < (1 << n); j++) {
    int s = 0;
    for (int i = 0; i < n; i++) {
        if (j & (1 << i)) {
            s += a[i];
        }
    }
}

高效枚举排列

 #include <algorithm>
 std::next_permutation 
 std::prev_permutation 

STL中的函数，接受两个参数表示数组的起止位置 会将数组变为对应的下/上一个排列 如果当前数组是最后一个排列，则变为第一个排列，并且返回false，否则返回true 时间复杂度(O(n))

int p[MAXN];
for (int i = 0; i < n; i++) p[i] = i + 1;
do {
    test(p);
} while (std::next_permutation(p, p + n));

免费送一道next_permutation 的裸题
P1088 火星人

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<string>
#include<cstring>
using namespace std;
inline int read() {
	char c = getchar(); int x = 0, f = 1;
	while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * f;
}
int a[100560],n,m;
signed main()
{
	cin>>n>>m;
	for(int i=0;i<n;++i) cin>>a[i];
	for(int i=1;i<=m;++i) next_permutation(a,a+n);
	for(int i=0;i<n;++i)
	{
		cout<<a[i]<<" ";
	}
	return 0;
}