AtCoder Beginner Contest 199（Sponsored by Panasonic） E - Permutation (状压dp)

• 题意:一长度为\(n\)的序列,有\(m\)个限制条件,问有多少排列方法使得题目所给的\(m\)个限制条件都满足.

• 题解:\(n\)给的范围很小,我们可以状态压缩,\(v[num][j]\)表示题目所给的限制条件前\(num\)个数最多不大于\(y\)的个数,我们可以枚举所有情况,然后判断每个状态是否和题目所给的条件冲突,如果冲突,我们就不转移.

• 代码:

  #include <bits/stdc++.h>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  #define rep(a,b,c) for(int a=b;a<=c;++a)
  #define per(a,b,c) for(int a=b;a>=c;--a)
  const int N = 1e3 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
  ll lcm(ll a,ll b) {return a/gcd(a,b)*b;}
   
  int n,m;
  int v[25][25];
  ll dp[1<<20];
   
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  	cin>>n>>m;
  	rep(i,0,20){
  		rep(j,0,20){
  			v[i][j]=INF;
  		}
  	}
   
  	rep(i,1,m){
  		int x,y,z;
  		cin>>x>>y>>z;
  		v[x][y-1]=min(v[x][y-1],z);
  	}
   
  	dp[0]=1;
  	rep(i,1,(1<<n)-1){
  		int num=0;
  		rep(j,0,n-1) if(i&(1<<j)) num++;
  		int sum=0;
  		bool flag=true;
  		rep(j,0,n-1){
  			if(i&(1<<j)) sum++;
  			if(sum>v[num][j]){
  				flag=false;
  				break;
  			}
  		}
  		if(flag){
  			rep(j,0,n-1){
  				if(i&(1<<j)){
  					dp[i]+=dp[i^(1<<j)];
  				}
  			}
  		}
  	}
   
  	cout<<dp[(1<<n)-1]<<'\n';
   
      return 0;
  }
  

标签：AtCoder,199,Beginner,int,ll,num,rep,dp,define
来源： https://www.cnblogs.com/lr599909928/p/14718619.html