【codeforces 798B】Mike and strings

【题目链接】:http://codeforces.com/contest/798/problem/B

【题意】

给你n个字符串;
每次操作,你可以把字符串的每个元素整体左移（最左边那个字符跑到最后面去了）
问你最少经过多少次操作可以使得所有字符串都相同;

【题解】

枚举最后每个字符串都变成了哪一个字符串;
然后每个字符串都模拟一下左移的过程;直到相等记录总的移动次数;
或者左移超过了长度的次数;输出不可能能和目标串一样;
记录最小的移动次数就好;

【Number Of WA】

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)typedef pair<int,int> pii;
typedef pair<LL,LL> pll;const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 110;
const int INF = 21e8;int n,ans=INF;
string s[N],s1[N];int ok(string t)
{rep1(i,1,n)s1[i]=s[i];int temp = 0,len = s1[1].size();rep1(i,1,n){int cnt = 0;while (s1[i]!=t){char t = s1[i][0];s1[i].erase(0,1);s1[i]+=t;cnt++;if (cnt>len+2) return INF;}temp+=cnt;}return temp;
}int main()
{//freopen("F:\\rush.txt","r",stdin);ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf�ͱ�����!cin >> n;rep1(i,1,n) cin >> s[i];rep1(i,1,n){int ju = ok(s[i]);ans = min(ans,ju);}if (ans>=INF)cout <<-1<<endl,0;elsecout << ans << endl;return 0;
}

转载于:https://www.cnblogs.com/AWCXV/p/7626396.html

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