Fun Game UVA - 1204

最新推荐文章于 2021-11-10 15:32:34 发布

AndrewThompson

最新推荐文章于 2021-11-10 15:32:34 发布

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分类专栏： Uva oj

本文链接：https://blog.youkuaiyun.com/zju2016/article/details/78100386

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本文介绍了一种优化的字符串匹配算法，用于解决特定场景下多个字符串组成的圆形图案问题。该算法通过预处理去除冗余字符串，并利用动态规划求解最优解，以减少重复字符的数量。

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首先读取所有的字符串，注意要保存所有字符串的读入顺序以及相应的反序（因为小朋友们维成的图形实际上是一个圆）。然后对所有读入的串按照长度进行排序，便于后续的处理，紧接着去除那些包含于其他某个串的串，最后就是计算串与串之间的重合的字符的个数。紧接着按照紫书的思路，从第一个串开始，后续的串加入到当前的集合当中，同时维护当前的除掉重合部分的长度，注意加入的同时既要加入原来的串也要加入原来串的反向串。最后还要处理结尾的串和起始串的重合部分，得出最终的结果。注意小孩子的个数最少为两个人。具体首先见如下代码：

#include<iostream>
#include<vector>
#include<string>
#include<set>
#include<stack>
#include<queue>
#include<map>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<deque>
using namespace std;

int N;

struct node{
	string fir, sec;
};

bool compare(const node a,const node b){
	return a.fir.length() < b.fir.length();
}

vector<node> record;
int repeat[20][2][20][2];
int dp[1 << 16][20][2];
vector<node> record2;

int cal_Rep(string a,string b){
	int n1 = a.size();
	int n2 = b.size();
	for (int i = 1; i < n1; i++){
		if (i + n2 <= n1) continue;
		bool suc = true;
		for (int j = 0; i + j < n1; j++){
			if (a[i + j] != b[j]){
				suc = false;
				break;
			}
		}
		if (suc) return n1 - i;
	}
	return 0;
}

void simple(){
	int amount = record.size();
	for (int i = 0; i < amount; i++){
		bool need = true;
		for (int j = i + 1; j < amount; j++){
			if (record[j].fir.find(record[i].fir) != string::npos ||
				record[j].sec.find(record[i].sec) != string::npos){
				need = false;
				break;
			}
		}
		if (need){
			record2.push_back(record[i]);
		}
	}
	amount = record2.size();
	memset(repeat,0,sizeof(repeat));
	for (int i = 0; i < amount; i++){
		for (int j = 0; j < amount; j++){
				repeat[i][0][j][0] = cal_Rep(record2[i].fir, record2[j].fir);
				repeat[i][1][j][0] = cal_Rep(record2[i].sec, record2[j].fir);
				repeat[i][0][j][1] = cal_Rep(record2[i].fir, record2[j].sec);
				repeat[i][1][j][1] = cal_Rep(record2[i].sec, record2[j].sec);
		}
	}
}

void solve(){
	dp[1][0][0] = record2[0].fir.size();
	int amount = record2.size();
	N = amount;
	for (int s = 1; s < (1 << N) - 1; s++){
		for (int i = 0; i < amount; i++){
			for (int j = 0; j < 2; j++){
				if (dp[s][i][j] != -1){
					for (int k = 0; k < amount; k++){
						if (!(s&(1 << k))){
							for (int t = 0; t < 2; t++){
								if (dp[s | (1 << k)][k][t] == -1)
									dp[s | (1 << k)][k][t] = dp[s][i][j] + record2[k].fir.size() - repeat[i][j][k][t];
								else
								dp[s | (1 << k)][k][t] =
									min(dp[s | (1 << k)][k][t], 
									(int)(dp[s][i][j] + record2[k].fir.size() - repeat[i][j][k][t]));
							}
						}
					}
				}
			}
		}
	}
	int res = -1;
	int total = (1 << N) - 1;
	for (int i = 0; i < amount; i++){
		for (int j = 0; j < 2; j++){
			if (dp[total][i][j] != -1){
				if (res == -1) res = dp[total][i][j] - repeat[i][j][0][0];
				else
					res = min(res, dp[total][i][j] - repeat[i][j][0][0]);
			}
		}
	}
	if (res <=1 ) res = 2;
	cout << res << endl;
}

int main(){
	while (cin >> N&&N){
		record.clear();
		record2.clear();
		for (int i = 0; i < N; i++){
			node temp;
			cin >> temp.fir;
			temp.sec = temp.fir;
			reverse(temp.sec.begin(), temp.sec.end());
			record.push_back(temp);
		}
		sort(record.begin(),record.end(),compare);
		memset(dp, -1, sizeof(dp));
		simple();
		solve();
	}
	return 0;
}