LA_3602DNA Consensus String ( hash, 貪心 )

題意：給定m個長度爲n的DNA序列，求一個DNA序列到所有DNA序列的Hamming距離最小，兩個字符串的Hamming距離定義爲：字符不同的位置數分析：統計每一列的ACGT的數量，取最多的爲rst，然後統計Hamming距離Code:#include <set>#include <map>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <deque>#include <vector>#include <cstdio>#include <bitset>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;#define DIR     4#define DIM     2#define STATUS  2#define MAXCNT  50 + 10#define MAXN    1000 + 10#define oo      (~0u)>>1#define INF     0x3F3F3F3F#define REPI(i, s, e)   for(int i = s; i <= e; i ++)#define REPD(i, e, s)   for(int i = e; i >= s; i --)static const double EPS = 1e-5;int hash[MAXCNT];char str[MAXCNT][MAXN], rst[MAXN];int main(int argc, char const *argv[]){#ifndef ONLINE_JUDGE        freopen("test.in", "r", stdin);#endif        int n, len, cas;        scanf("%d", &cas);        REPI(k, 1, cas) {                scanf("%d %d", &n, &len);                REPI(i, 0, n-1) {                        scanf("%s", str[i]);                }                int ans = 0;                REPI(c, 0, len-1) {                        hash['A'-'A'] = hash['C'-'A'] = 0;                        hash['T'-'A'] = hash['G'-'A'] = 0;                        REPI(r, 0, n-1) {                                hash[str[r][c]-'A'] += 1;                        }                        char ch = 'A';                        int max_val = hash['A'-'A'], sum = 0;                        sum += hash['A'-'A'];                        sum += hash['C'-'A'];                        sum += hash['G'-'A'];                        sum += hash['T'-'A'];                        if( max_val < hash['C'-'A'] ) {                                max_val = hash['C'-'A'], ch = 'C';                        }                        if( max_val < hash['G'-'A'] ) {                                max_val = hash['G'-'A'], ch = 'G';                        }                        if( max_val < hash['T'-'A'] ) {                                max_val = hash['T'-'A'], ch = 'T';                        }                        rst[c] = ch, ans += sum-hash[ch-'A'];                }                REPI(i, 0, len-1) {                        printf("%c", rst[i]);                }                printf("\n%d\n", ans);        }        return 0;}