PAT (Top Level) Practice1004 To Buy or Not to Buy - Hard Version （35 分）

最新推荐文章于 2020-03-31 10:53:22 发布

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本文深入探讨了深度优先搜索（DFS）算法在解决特定问题时的应用与优化技巧，通过实例展示了如何结合剪枝策略减少不必要的搜索，提高算法效率。文章详细介绍了在字符串匹配和路径寻找等问题中，如何利用DFS进行有效求解，并给出了具体实现代码。

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题目链接：https://pintia.cn/problem-sets/994805148990160896/problems/994805155206119424

暴力，基本上不会超时，dfs加上一些剪枝

算法：dfs

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#define N 150
#define inf 0x3f3f3f
using namespace std;
int n;
int op[N][N];
int po[N][N];
string str[N];
string q[N];
int num[N];
int ans = 0;
int vis[N];
int re = inf;
int flagss = 0;
void dfs(int index,int u,string s) {
	if (flagss == 1)
		return;
	if (s.size() == 0) {
		re = min(re, index);
		if (re == 0) {
			flagss = 1;
		}
		return ;
	}
	for (int i = u; i <= ans; i++) {
		set<char>p;
		int ks[N];
		int sd = 0;
		for (int j = 0; j < s.size();j++) {
			if (!p.count(s[j])) {
				if (op[i][s[j]] == 0)
				{
					sd = 1;
					break;
				}
				ks[s[j]] = 1;
				p.insert(s[j]);
			}
			else {
				if (ks[s[j]] + 1 > op[i][s[j]]) {
					sd = 1;
					break;
				}
				ks[s[j]]++;
			}
		}
		if (sd == 1)
		{
			break;
		}		
		string s1 = s;
		int nu1 = str[num[i]].size();
		int nu2 = s1.size();
		for (string::iterator iter = q[i].begin(); iter != q[i].end();) {
			if (s1.find_first_of(*iter) != -1)
			{
				s1.erase(s1.find_first_of(*iter), 1);
				iter++;
			}
			else {
				iter++;
			}
		}
		int nu = nu1 - (nu2 - s1.size());
		dfs(index + nu, u + 1, s1);
		if (flagss == 1)
			return;
	}
}
int main() {
	string s;
	string s2;
	cin >> s;
	s2 = s;
	memset(vis, 0, sizeof(vis));
	scanf("%d", &n);
	int flag = 0;
	int flags = 0;
	for (int i = 0; i < n; i++) {
		cin >> str[i];
		if (str[i] == s)
		{
			flags = 1;
		}
		if (flags == 1)
			continue;
		flag = 0;
		for (int j = 0; j < str[i].size(); j++) {
			if (s.find(str[i][j])!=string::npos) {
				if (flag == 0) {
					q[++ans] = "";
					q[ans] += str[i][j];
					po[ans][str[i][j]] = 1;
					num[ans] = i;
					flag = 1;
				}
				else {
					q[ans] += str[i][j];
					po[ans][str[i][j]]++;
				}
			}
		}
	}
	if (flags == 1)
	{
		cout << "Yes " << flags-1 << endl;
		return 0;
	}
	memset(op, 0, sizeof(op));
	for (int i = ans; i >= 1; i--) {
		for (int j = 0; j < s.size(); j++) {
			if (q[i].find(s[j]) != string::npos) {
				    op[i][s[j]] = op[i + 1][s[j]] + po[i][s[j]];
			}
			else {
				op[i][s[j]] = op[i + 1][s[j]];
			}
		}
	}
	dfs(0, 1,s);
	if (re == inf)
	{
		for (int i = 1; i <= ans; i++) {
			for (string::iterator iter = q[i].begin(); iter != q[i].end();) {
				if (s2.find_first_of(*iter) != -1)
				{
					s2.erase(s2.find_first_of(*iter), 1);
					iter++;
				}
				else {
					iter++;
				}
			}
		}
		cout << "No " <<s2.size()<< endl;
	}
	else
	    cout <<"Yes " <<re<< endl;
	return 0;
}