Codeforces_442_A

Posted 2020-08-09 冷暖知不知

http://codeforces.com/problemset/problem/442/A

 

想想成5*5的图，一共能划10条线，枚举2^10次即可。

判断每种情况是否符合条件的方法，若存在点，被线穿过的状态一样，当时不是同一个点，则不符合。

 

#include <cstdio>
#include<iostream>
#include<string>
#include<cstring>
using namespace std;
int color[105],num[105],temp[105],count[1050] = {0};
string s;
int main()
{
    int n;
    cin >> n;
    for(int i = 1;i <= n;i++)
    {
        cin >> s;
        if(s[0] == ‘R‘) color[i] = 0;
        else if(s[0] == ‘G‘) color[i] = 1;
        else if(s[0] == ‘B‘) color[i] = 2;
        else if(s[0] == ‘Y‘) color[i] = 3;
        else if(s[0] == ‘W‘) color[i] = 4;
        num[i] = s[1]-‘1‘;
    }
    int ans = 10;
    for(int i = 0;i < (1<<10);i++)
    {
        memset(temp,0,sizeof(temp));
        if(i)
            count[i] = count[i>>1]+(i&1);
        for(int j = 1;j <= n;j++)
        {
            if((i>>color[j])&1)
                temp[j] |= 1<<color[j];
            if((i>>(num[j]+5))&1)
                temp[j] |= 1<<(num[j]+5);
            for(int k = 1;k < j;k++)
            {
                if(temp[j] == temp[k] && (color[k] != color[j] || num[j] != num[k]))
                    goto there;
            }
        }
        ans = min(ans,count[i]);
        there:
        continue;
    }
    cout << ans << endl;
    return 0;
}