【Codeforces】1042 - B Vitamins

最新推荐文章于 2020-05-04 20:03:14 发布

Dicer_

最新推荐文章于 2020-05-04 20:03:14 发布

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CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/qq_41009682/article/details/82794560

本文介绍了一种使用二进制枚举优化动态规划算法的方法，通过具体实例展示了如何利用二进制特性减少状态空间，从而提高算法效率。文章详细解释了代码实现过程，包括初始化状态、转移方程及最终求解。

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题目链接

先看一下我暴力的代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define clr(s, x) memset(s, x, sizeof(s))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
inline int read(){int r=0;char c=getchar();while(c<'0'||c>'9') {c=getchar();}while(c>='0'&&c<='9') {r=r*10+c-'0';c=getchar();}return r;}
inline ll readll(){ll r=0;char c=getchar();while(c<'0'||c>'9') {c=getchar();}while(c>='0'&&c<='9') {r=r*10+c-'0';c=getchar();}return r;}
inline ll qpow(ll a,ll b,ll mod){ll res=1;while(b){if(b&1)res = (res*a)%mod;a=(a*a)%mod;b>>=1;}return res;}
inline ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const double eps = 1e-8;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
const int MAXN = 1e5;
const int MAXM = 1e5;

int dp[10], p;
void solve(int res){
		if(res == 1){
			dp[1] = min(dp[1], dp[0] + p);
			dp[3] = min(dp[3], dp[2] + p);
			dp[5] = min(dp[5], dp[4] + p);
			dp[7] = min(dp[7], dp[6] + p);
		} else if(res == 2){
			dp[2] = min(dp[2], dp[0] + p);
			dp[3] = min(dp[3], dp[1] + p);
			dp[6] = min(dp[6], dp[4] + p);
			dp[7] = min(dp[7], dp[5] + p);
		} else if(res == 3){
			dp[3] = min(dp[3], dp[0] + p);
			dp[7] = min(dp[7], dp[4] + p);
		} else if(res == 4){
			dp[4] = min(dp[4], dp[0] + p);
			dp[5] = min(dp[5], dp[1] + p);
			dp[6] = min(dp[6], dp[2] + p);
			dp[7] = min(dp[7], dp[3] + p);
		} else if(res == 5){
			dp[5] = min(dp[5], dp[0] + p);
			dp[7] = min(dp[7], dp[2] + p);
		} else if(res == 6){
			dp[6] = min(dp[6], dp[0] + p);
			dp[7] = min(dp[7], dp[1] + p);
		} else if(res == 7){
			dp[7] = min(dp[7], dp[0] + p);
		}
}
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	int n;
	while(cin >> n){
		int res;
		string s;
		clr(dp, 0x3f);
		dp[0] = 0;
		for(int i=1; i<=n; ++i){
			cin >> p;
			cin >> s;
			res = 0;
			if((int)s.find('A') != -1)	res+=1;
			if((int)s.find('B') != -1)	res+=2;
			if((int)s.find('C') != -1)	res+=4;
			if(res == 1){
				solve(1);
			} else if(res == 2){
				solve(2);
			} else if(res == 3){
				solve(1);
				solve(2);
				solve(3);
			} else if(res == 4){
				solve(4);
			} else if(res == 5){
				solve(1);
				solve(4);
				solve(5);
			} else if(res == 6){
				solve(2);
				solve(4);
				solve(6);
			} else if(res == 7){
				solve(1);
				solve(2);
				solve(3);
				solve(4);
				solve(5);
				solve(6);
				solve(7);
			}
			// cout << dp[7] << endl;
		}
		if(dp[7] >= INF)	puts("-1");
		else cout << dp[7] << endl;
	}
	return 0;
}

看完应该就知道思路了，然后我们用二进制枚举来做：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define clr(s, x) memset(s, x, sizeof(s))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
inline int read(){int r=0;char c=getchar();while(c<'0'||c>'9') {c=getchar();}while(c>='0'&&c<='9') {r=r*10+c-'0';c=getchar();}return r;}
inline ll readll(){ll r=0;char c=getchar();while(c<'0'||c>'9') {c=getchar();}while(c>='0'&&c<='9') {r=r*10+c-'0';c=getchar();}return r;}
inline ll qpow(ll a,ll b,ll mod){ll res=1;while(b){if(b&1)res = (res*a)%mod;a=(a*a)%mod;b>>=1;}return res;}
inline ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const double eps = 1e-8;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
const int MAXN = 1e5;
const int MAXM = 1e5;

int dp[10], p;
void solve(int res){
	for(int i=1; i<=7; ++i){
		bool flag = 1;
		for(int j=0; j<=2; ++j){
			if(((res>>j)&1) && !((i>>j)&1))	flag = 0;
		}
		if(flag) dp[i] = min(dp[i], dp[i-res] + p);
	}
}
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	int n;
	while(cin >> n){
		int res;
		string s;
		clr(dp, 0x3f);
		dp[0] = 0;
		for(int i=1; i<=n; ++i){
			cin >> p;
			cin >> s;
			res = 0;
			if((int)s.find('A') != -1)	res+=1;
			if((int)s.find('B') != -1)	res+=2;
			if((int)s.find('C') != -1)	res+=4;
			// cout << res << endl;
			for(int i=1; i<=7; ++i){
				bool flag = 1;
				for(int j=0; j<=2; ++j){
					if(((i>>j)&1) && !((res>>j)&1)){
						flag = 0;
					}
				}
				if(flag) solve(i);
			}
			// cout << dp[7] << endl;
		}
		if(dp[7] >= INF)	puts("-1");
		else cout << dp[7] << endl;
	}
	return 0;
}

就酱~