【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

0

【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;
    else
        cout << ans << endl;
    return 0;
}
原文地址:https://www.cnblogs.com/AWCXV/p/7626396.html