UVA-1368 DNA Consensus String

原题地址

G - DNA Consensus String

解题思路

直接对每一列在线处理，更新当前出现次数最多的字符，最后将每列的个数减去该字符的数量，就得到与之不同的字符个数。累加得到题目要求的Hamming distances。

参考代码

#include<bits/stdc++.h>
using namespace std;
#define LOCAL  //提交的时候一定注释
#define _for(i, a, b) for(int i = (a); i < (b); ++i)
#define _rep(i, a, b) for(int i = (a); i <= (b); ++i)
#define pb push_back
#define VI vector<int>
#define maxn 30
#define INF 0x3f3f3f3f
typedef long long LL;
vector<string> str;
int num[maxn];

int readint() {
    
    
    int x; scanf("%d", &x); return x;
}

int main() {
    
    
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
#endif
    int t, n, m;
    scanf("%d", &t);
    string s;
    while (t--) {
    
    
        str.clear();
        scanf("%d%d\n", &n, &m);
        _for(i, 0, n) {
    
    
            getline(cin, s);
            str.pb(s);
        }
        int sum = 0, less = 0;
        string ans = "";
        _for(j, 0, m) {
    
    
            memset(num, 0, sizeof(num));
            int cnt = 0;
            char now = 'Z', tmp;
            _for(i, 0, n) {
    
    
                tmp = str[i][j];
                num[tmp - 'A']++;
                if (num[tmp - 'A'] > cnt || (num[tmp - 'A'] == cnt && tmp < now)) {
    
    
                    now = tmp;
                    cnt = num[tmp - 'A'];
                }
            }
            ans = ans + now;
            less += (n - num[now - 'A']);
        }
        printf("%s\n%d\n", ans.c_str(), less);
    }
    return 0;
}