LA 3602 DNA Consensus String

题意：给m个长度为n的字符串，求一个长度为n的字符串，使得它和该m个不同字符之和最小，只有A、T、C、G四种字符

题解：直接暴力贪心，对于每一列，取最多的那个字符即可

 

#include<bits/stdc++.h>
using namespace std;
char s[52][1005];
int main()
{
        ios::sync_with_stdio(false);
        int test;
        cin >> test;
        while( test -- )
        {
                char t[1000+5];
                int n,m;
                cin >> n >> m;
                for(int i = 1;i <= n; i++ ) cin >> s[i]+1;
                int ans = 0;
                int pos[27];
                for(int j = 1;j <= m; j++)
                {
                        memset(pos,0,sizeof(pos));
                        for(int i = 1;i <= n;i++) pos[s[i][j]-'A']++;
                        int sum = max(pos[0],max(pos[2], max( pos[6], pos[('T'-'A')] ) ));
                        ans += (n-sum);
                        if(pos[0] >= pos[2] && pos[0] >= pos[6] && pos[0] >= pos['T'-'A']) t[j]='A';
                        else if( pos[0] < pos[2] && pos[2] >= pos[6] && pos[2] >= pos['T'-'A'] ) t[j] = 'C';
                        else if(pos[6] >= pos['T'-'A'] && pos[6] > pos[0] && pos[6] > pos[2]) t[j] = 'G';
                        else t[j] = 'T';
                }
                t[m+1] = '\0';
                cout << t+1 << endl;
                cout << ans << endl;
        }
}