习题7-5(uva-690)

#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int maxn = 35;
int n, origin[5], moveStep[maxn], moveCnt,ans;

//要让大爷有位置
bool CanAddHead(int s0,int s1,int s2,int s3,int s4) {
	return ((s0 & origin[0]) == 0) &&
		((s1 & origin[1]) == 0) &&
		((s2 & origin[2]) == 0) &&
		((s3 & origin[3]) == 0) &&
		((s4 & origin[4]) == 0);
}
void Init()
{
	string input;
	moveCnt = 0;
	ans = n * 10;
	for (int i = 0; i < 5; i++) {
		cin >> input;
		origin[i] = 0;
		//把 大爷 的情况记录下来 每次来个大爷都得给他让位置
		for (int j = input.size() - 1; j >= 0; j--) {
			origin[i] = (origin[i] << 1) + (input[j] == 'X');
		}
	}

	for (int i = 0; i <= n; i++) {
		//前面的挤一挤 后面来人能不能有位置
		if (CanAddHead(origin[0]>>i,origin[1]>>i,origin[2]>>i,origin[3]>>i,origin[4]>>i)) moveStep[moveCnt++] = i;
	}
}

void dfs(int s0, int s1, int s2, int s3, int s4, int d, int len) {
	//
	if (len + moveStep[0] * (10 - d) > ans) return;
	//弄完了
	if (d == 10) {
		ans = min(ans, len);
		return;
	}

	for (int i = 0; i < moveCnt; i++) {
		int rCnt = moveStep[i];
		//让位置给大爷 大爷
		int ss0 = s0 >> rCnt, ss1 = s1 >> rCnt, ss2 = s2 >> rCnt, ss3 = s3 >> rCnt, ss4 = s4 >> rCnt;
		//大爷能不能坐下 前面的挤下去没影响 最多只影响20
		if (CanAddHead(ss0, ss1, ss2, ss3, ss4)) {
			dfs(ss0 | origin[0], ss1 | origin[1], ss2 | origin[2], ss3 | origin[3], ss4 | origin[4],d + 1,len + rCnt);
		}
	}
}
int main()
{
	while (cin >> n && n) {
		Init();
		dfs(origin[0], origin[1], origin[2], origin[3], origin[4], 1, n);
		cout << ans << endl;
	}
	return 0;
}